Evaluate $sumlimits_{n=0}^{infty}(-1)^nsumlimits_{j=0}^{k}{k choose j}frac{(-1)^j}{2n+2j+1}$
$begingroup$
Evaluate $g(k)$ if $sumlimits_{n=0}^{infty}(-1)^nsumlimits_{j=0}^{k}{k choose j}frac{(-1)^j}{2n+2j+1}=frac{pi}{2^{2-k}}+g(k)$.
The well-known Gregory Series,
$$sum_{n=0}^{infty}frac{(-1)^n}{2n+1}=frac{pi}{4}tag1$$
Let us generalize $(1)$ in terms of binomial coefficients ${k choose j}$,
Where $kge0$
$$sum_{n=0}^{infty}(-1)^nsum_{j=0}^{k}{k choose j}frac{(-1)^j}{2n+2j+1}=frac{pi}{2^{2-k}}+g(k)tag2$$
Expand $(2)$ for $k=1,2$ and $3$,
$$sum_{n=0}^{infty}(-1)^nleft(frac{1}{2n+1}-frac{1}{2n+3}right)=frac{pi}{2}-1tag3$$
$$sum_{n=0}^{infty}(-1)^nleft(frac{1}{2n+1}-2frac{1}{2n+3}+frac{1}{2n+5}right)=pi-frac{8}{3}tag4$$
$$sum_{n=0}^{infty}(-1)^nleft(frac{1}{2n+1}-frac{3}{2n+3}+frac{3}{2n+5}-frac{1}{2n+5}right)=2pi-frac{88}{15}tag5$$
The pattern for $g(k)$ is not so obvious, I was not able to determine the closed form $g(k)$
Q:
How can we find the closed form for $(2)?$
sequences-and-series
edited Jan 1 at 17:35
Did
249k23229468
asked Dec 31 '18 at 16:32
EndgameEndgame
1
$endgroup$
add a comment |
$begingroup$
Evaluate $g(k)$ if $sumlimits_{n=0}^{infty}(-1)^nsumlimits_{j=0}^{k}{k choose j}frac{(-1)^j}{2n+2j+1}=frac{pi}{2^{2-k}}+g(k)$.
The well-known Gregory Series,
$$sum_{n=0}^{infty}frac{(-1)^n}{2n+1}=frac{pi}{4}tag1$$
Let us generalize $(1)$ in terms of binomial coefficients ${k choose j}$,
Where $kge0$
$$sum_{n=0}^{infty}(-1)^nsum_{j=0}^{k}{k choose j}frac{(-1)^j}{2n+2j+1}=frac{pi}{2^{2-k}}+g(k)tag2$$
Expand $(2)$ for $k=1,2$ and $3$,
$$sum_{n=0}^{infty}(-1)^nleft(frac{1}{2n+1}-frac{1}{2n+3}right)=frac{pi}{2}-1tag3$$
$$sum_{n=0}^{infty}(-1)^nleft(frac{1}{2n+1}-2frac{1}{2n+3}+frac{1}{2n+5}right)=pi-frac{8}{3}tag4$$
$$sum_{n=0}^{infty}(-1)^nleft(frac{1}{2n+1}-frac{3}{2n+3}+frac{3}{2n+5}-frac{1}{2n+5}right)=2pi-frac{88}{15}tag5$$
The pattern for $g(k)$ is not so obvious, I was not able to determine the closed form $g(k)$
Q:
How can we find the closed form for $(2)?$
sequences-and-series
edited Jan 1 at 17:35
Did
249k23229468
asked Dec 31 '18 at 16:32
EndgameEndgame
1
$endgroup$
add a comment |
$begingroup$
Evaluate $g(k)$ if $sumlimits_{n=0}^{infty}(-1)^nsumlimits_{j=0}^{k}{k choose j}frac{(-1)^j}{2n+2j+1}=frac{pi}{2^{2-k}}+g(k)$.
The well-known Gregory Series,
$$sum_{n=0}^{infty}frac{(-1)^n}{2n+1}=frac{pi}{4}tag1$$
Let us generalize $(1)$ in terms of binomial coefficients ${k choose j}$,
Where $kge0$
$$sum_{n=0}^{infty}(-1)^nsum_{j=0}^{k}{k choose j}frac{(-1)^j}{2n+2j+1}=frac{pi}{2^{2-k}}+g(k)tag2$$
Expand $(2)$ for $k=1,2$ and $3$,
$$sum_{n=0}^{infty}(-1)^nleft(frac{1}{2n+1}-frac{1}{2n+3}right)=frac{pi}{2}-1tag3$$
$$sum_{n=0}^{infty}(-1)^nleft(frac{1}{2n+1}-2frac{1}{2n+3}+frac{1}{2n+5}right)=pi-frac{8}{3}tag4$$
$$sum_{n=0}^{infty}(-1)^nleft(frac{1}{2n+1}-frac{3}{2n+3}+frac{3}{2n+5}-frac{1}{2n+5}right)=2pi-frac{88}{15}tag5$$
The pattern for $g(k)$ is not so obvious, I was not able to determine the closed form $g(k)$
Q:
How can we find the closed form for $(2)?$
sequences-and-series
edited Jan 1 at 17:35
Did
249k23229468
asked Dec 31 '18 at 16:32
EndgameEndgame
1
$endgroup$
Evaluate $g(k)$ if $sumlimits_{n=0}^{infty}(-1)^nsumlimits_{j=0}^{k}{k choose j}frac{(-1)^j}{2n+2j+1}=frac{pi}{2^{2-k}}+g(k)$.
The well-known Gregory Series,
$$sum_{n=0}^{infty}frac{(-1)^n}{2n+1}=frac{pi}{4}tag1$$
Let us generalize $(1)$ in terms of binomial coefficients ${k choose j}$,
Where $kge0$
$$sum_{n=0}^{infty}(-1)^nsum_{j=0}^{k}{k choose j}frac{(-1)^j}{2n+2j+1}=frac{pi}{2^{2-k}}+g(k)tag2$$
Expand $(2)$ for $k=1,2$ and $3$,
$$sum_{n=0}^{infty}(-1)^nleft(frac{1}{2n+1}-frac{1}{2n+3}right)=frac{pi}{2}-1tag3$$
$$sum_{n=0}^{infty}(-1)^nleft(frac{1}{2n+1}-2frac{1}{2n+3}+frac{1}{2n+5}right)=pi-frac{8}{3}tag4$$
$$sum_{n=0}^{infty}(-1)^nleft(frac{1}{2n+1}-frac{3}{2n+3}+frac{3}{2n+5}-frac{1}{2n+5}right)=2pi-frac{88}{15}tag5$$
The pattern for $g(k)$ is not so obvious, I was not able to determine the closed form $g(k)$
Q:
How can we find the closed form for $(2)?$
sequences-and-series
sequences-and-series
edited Jan 1 at 17:35
Did
249k23229468
asked Dec 31 '18 at 16:32
EndgameEndgame
1
edited Jan 1 at 17:35
Did
249k23229468
asked Dec 31 '18 at 16:32
EndgameEndgame
1
edited Jan 1 at 17:35
Did
249k23229468
edited Jan 1 at 17:35
Did
249k23229468
edited Jan 1 at 17:35
Did
249k23229468
249k23229468
asked Dec 31 '18 at 16:32
EndgameEndgame
1
asked Dec 31 '18 at 16:32
EndgameEndgame
1
asked Dec 31 '18 at 16:32
EndgameEndgame
1
1
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
We will make use of the identity
$$
sum_{k=0}^n(-1)^kbinom{n}{k}frac1{k+alpha}=frac{n!,Gamma(alpha)}{Gamma(n+1+alpha)}tag1
$$
which can be proven by induction on $n$.
The sum in the question is
$$
begin{align}
&sum_{n=0}^infty(-1)^nsum_{j=0}^kbinom{k}{j}frac{(-1)^j}{2n+2j+1}tag2\
&=frac12sum_{n=0}^infty(-1)^nsum_{j=0}^kbinom{k}{j}frac{(-1)^j}{n+j+frac12}tag3\
&=frac12sum_{n=0}^infty(-1)^nfrac{k!,Gamma!left(n+frac12right)}{Gamma!left(n+k+frac32right)}tag4\
&=frac12sum_{n=0}^infty(-1)^nint_0^1(1-t)^kt^{n-frac12},mathrm{d}ttag5\
&=frac12int_0^1frac{(1-t)^k}{(1+t)sqrt{t}},mathrm{d}ttag6\
&=int_0^1frac{left(1-t^2right)^k}{1+t^2},mathrm{d}ttag7\
&=int_0^1frac{left(2-left(1+t^2right)right)^k}{1+t^2},mathrm{d}ttag8\
&=2^kleft(fracpi4+sum_{j=1}^kleft(-frac12right)^jbinom{k}{j}int_0^1left(1+t^2right)^{j-1},mathrm{d}tright)tag9\
&=2^kleft(fracpi4+sum_{j=1}^kleft(-frac12right)^jbinom{k}{j}sum_{i=0}^{j-1}binom{j-1}{i}frac1{2i+1}right)tag{10}\
&=2^kleft(fracpi4-frac12sum_{j=0}^{k-1}left(-frac12right)^jsum_{m=1}^kbinom{k-m}{j}sum_{i=0}^{j}binom{j}{i}frac1{2i+1}right)tag{11}\
&=2^kleft(fracpi4-frac12sum_{m=1}^ksum_{i=0}^{k-1}sum_{j=i}^{k-m}left(-frac12right)^jbinom{k-m}{i}binom{k-m-i}{j-i}frac1{2i+1}right)tag{12}\
&=2^kleft(fracpi4-frac12sum_{m=1}^ksum_{i=0}^{k-1}sum_{j=0}^{k-m-i}left(-frac12right)^{j+i}binom{k-m}{i}binom{k-m-i}{j}frac1{2i+1}right)tag{13}\
&=2^kleft(fracpi4-frac12sum_{m=1}^ksum_{i=0}^{k-1}left(-frac12right)^ibinom{k-m}{i}left(frac12right)^{k-m-i}frac1{2i+1}right)tag{14}\
&=2^kleft(fracpi4-frac14sum_{m=1}^ksum_{i=0}^{k-m}(-1)^ibinom{k-m}{i}left(frac12right)^{k-m}frac1{i+frac12}right)tag{15}\
&=2^kleft(fracpi4-frac14sum_{m=1}^kleft(frac12right)^{k-m}frac{(k-m)!,Gamma!left(frac12right)}{Gamma!left(k-m+frac32right)}right)tag{16}\
&=2^kleft(fracpi4-frac14sum_{m=0}^{k-1}left(frac12right)^mfrac{m!,Gamma!left(frac12right)}{Gamma!left(m+frac32right)}right)tag{17}\
&=2^kleft(fracpi4-frac12sum_{m=0}^{k-1}frac{m!}{(2m+1)!!}right)tag{18}\
end{align}
$$
Explanation:
$phantom{1}(3)$: bring $frac12$ out front
$phantom{1}(4)$: apply $(1)$
$phantom{1}(5)$: apply the Beta Function
$phantom{1}(6)$: sum in $n$
$phantom{1}(7)$: substitute $tmapsto t^2$
$phantom{1}(8)$: $1-t^2=2-left(1+t^2right)$
$phantom{1}(9)$: apply the Binomial Theorem
$(10)$: apply the Binomial Theorem and integrate
$(11)$: $binom{k}{j}=sumlimits_{m=1}^kbinom{k-m}{j-1}$ and substitute $jmapsto j+1$
$(12)$: reorder summation and $binom{k-m}{j}binom{j}{i}=binom{k-m}{i}binom{k-m-i}{j-i}$
$(13)$: substitute $jmapsto j+i$
$(14)$: apply the Binomial Theorem
$(15)$: shift factors of $2$
$(16)$: apply $(1)$
$(17)$: substitute $mmapsto k-m$
$(18)$: rewrite the Gamma functions as factorials
Thus,
$$
bbox[5px,border:2px solid #C0A000]{g(k)=-2^{k-1}sum_{m=0}^{k-1}frac{m!}{(2m+1)!!}}tag{19}
$$
$(4)$ says that asymptotically, the full sum from the question is $frac12sqrt{fracpi{k}}$, which vanishes as $ktoinfty$. Thus, the sum in $(18)$ must converge quickly to $fracpi2$. In fact, the sum in $(18)$ is the partial sum of the sum in this question whose limit is $fracpi2$.
answered Jan 1 at 14:22
robjohn♦robjohn
271k27317644
$endgroup$
add a comment |
$begingroup$
Note that the partial fractions decomposition of the inverse
of the Rising Factorial gives:
$$
eqalign{
& left( {x - 1} right)^{,underline {, - left( {m + 1} right),} } = {1 over {x^{,overline {,m + 1,} } }}
= {{Gamma (x)} over {Gamma (x + m + 1)}} = ;quad left| {;0 le {rm integer }, m} right.quad = cr
& = sumlimits_{0, le ,k, le ,m} {{1 over {x + k}}left( {{{left( { - 1} right)^{,k} } over {left( {m - k} right)!k!}}} right)}
= sumlimits_{0, le ,k, le ,m} {{1 over {x + k}}left( {{{left( { - 1} right)^{,k} } over {m!}}binom{m}{k}
} right)} cr}
$$
So
$$
eqalign{
& sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
{1 over {2n + 2j + 1}}} = cr
& = {1 over 2}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
{1 over {left( {n + 1/2 + j} right)}}} = cr
& = {{k!} over {2left( {n + 1/2} right)^{,overline {,k + 1,} } }} cr}
$$
We have then that the sum in $n$ is
$$
eqalign{
& S(k) = sumlimits_{0, le ,n} {left( { - 1} right)^{,n} sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right){1 over {2n + 2j + 1}}} } = cr
& = {1 over 2}sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{k!} over {left( {n + 1/2} right)^{,overline {,k + 1,} } }}} = cr
& = {1 over {2left( {k + 1} right)}}sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{1^{,overline {,k + 1,} } }
over {left( {1 + n - 1/2} right)^{,overline {,k + 1,} } }}} = cr
& = {1 over {2left( {k + 1} right)}}sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{1^{,overline {,n - 1/2,} } }
over {left( {2 + k} right)^{,overline {,n - 1/2,} } }}} = cr
& = {{1^{,overline {, - 1/2,} } } over {2left( {k + 1} right)left( {2 + k} right)^{,overline {, - 1/2,} } }}
sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{left( {1/2} right)^{,overline {,n,} } } over {left( {3/2 + k} right)^{,overline {,n,} } }}} = cr
& = {{Gamma left( {1/2} right)Gamma left( {2 + k} right)} over {2left( {k + 1} right)Gamma left( 1 right)Gamma left( {3/2 + k} right) }}
sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{left( {1/2} right)^{,overline {,n,} } } over {left( {3/2 + k} right)^{,overline {,n,} } }}} = cr
& = {{Gamma left( {1/2} right)Gamma left( {1 + k} right)} over {2,Gamma left( {3/2 + k} right)}}{}_2F_{,1} left( {left. {matrix{
{1/2;,;1} cr
{k + 3/2} cr
} ;} right| - 1} right) = cr
& = 2^{,2,k} {{left( {k!} right)^{,2} } over {left( {2k + 1} right)!}};;{}_2F_{,1} left( {left. {matrix{
{1/2;,;1} cr
{k + 3/2} cr
} ;} right| - 1} right) cr}
$$
where in the mid of the derivation we have used the identity
$$
{{z^{,overline {,w,} } } over {left( {z + a} right)^{,overline {,w,} } }} = {{z^{,overline {,a,} } } over {left( {z + w} right)^{,overline {,a,} } }}
$$
and at the end the duplication formula for Gamma
$$
Gamma left( {2,z} right) = {{2^{,2,z - 1} } over {sqrt pi }}Gamma left( z right)Gamma left( {z + 1/2} right) = 2^{,2,z - 1} {{Gamma left( z right)Gamma left( {z + 1/2} right)} over {Gamma left( {1/2} right)}}
$$
Alternative Approach
Actually there is a more straight alternative aproach through the Digamma Function
$$
eqalign{
& S(k) = sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {1 over {2n + 2j + 1}}} } = cr
& = sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)sumlimits_{0, le ,m} {left( {{1 over {4m + 2j + 1}} - {1 over {4m + 2j + 3}}} right)} } = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)sumlimits_{0, le ,m} {left( {{1 over {m + j/2 + 1/4}} - {1 over {m + j/2 + 3/4}}} right)} } = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)left( {psi left( {j/2 + 3/4} right) - psi left( {j/2 + 1/4} right)} right)} = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,} {left( matrix{
k cr
2j cr} right)left( {psi left( {j + 3/4} right) - psi left( {j + 1/4} right)} right) - left( matrix{
k cr
2j + 1 cr} right)left( {psi left( {j + 5/4} right) - psi left( {j + 3/4} right)} right)} = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,} {left( matrix{
k + 1 cr
2j + 1 cr} right)psi left( {j + 3/4} right) - left( matrix{
k cr
2j cr} right)psi left( {j + 1/4} right) - left( matrix{
k cr
2j + 1 cr} right)psi left( {j + 1 + 1/4} right)} = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,} {left( matrix{
k + 1 cr
2j + 1 cr} right)left( {psi left( {j + 3/4} right) - psi left( {j + 1/4} right)} right) - left( matrix{
k cr
2j + 1 cr} right){1 over {j + 1/4}}} cr}
$$
Since we can write (re. for instance to this page)
$$
eqalign{
& psi left( {j + 3/4} right) = 4sumlimits_{0, le ,l, le ,j - 1,} {{1 over {4,l + 3}}} + {pi over 2} - ln 8 - gamma
= psi left( {j + 3/4} right) - left( {psi left( {3/4} right) - left( {{pi over 2} - ln 8 - gamma } right)} right) cr
& psi left( {j + 1/4} right) = 4sumlimits_{0, le ,l, le ,j - 1,} {{1 over {4,l + 1}}} - {pi over 2} - ln 8 - gamma
= psi left( {j + 1/4} right) - left( {psi left( {1/4} right) - left( { - {pi over 2} - ln 8 - gamma } right)} right) cr}
$$
we arrive to express the sum as
$$
eqalign{
& S(k) = sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {1 over {2n + 2j + 1}}} } = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
left( {psi left( {j/2 + 3/4} right) - psi left( {j/2 + 1/4} right)} right)} = cr
& = {pi over 4}2^{,k} - 2sumlimits_{left( {0, le } right),j,left( { le ,k} right),} {binom{k+1}{2j+1}
left( {sumlimits_{0, le ,l, le ,j - 1,} {{1 over {left( {4,l + 3} right)left( {4,l + 1} right)}}} } right)}
- sumlimits_{left( {0, le } right),j,left( { le ,left( {k - 1} right)/2} right),} {binom{k}{2j+1}{1 over {4j + 1}}} = cr
& = {pi over 4}2^{,k} - 2sumlimits_{left( {0, le } right),j,left( { le ,k} right),} {binom{k+1}{2j+1}
left( {sumlimits_{0, le ,l, le ,j - 1,} {{1 over {left( {4,l + 2} right)^{,2} - 1}}} } right)}
- sumlimits_{left( {0, le } right),j,left( { le ,left( {k - 1} right)/2} right),} {binom{k}{2j+1}{1 over {4j + 1}}} cr}
$$
Interesting the comparison of this expression with the one given by
@robjohn
answered Jan 1 at 17:31
G CabG Cab
20.5k31342
$endgroup$
add a comment |
$begingroup$
(2) is
$$frac11-frac23+frac25-frac27+cdots=-1+2left(frac11-frac13+frac15-cdotsright),$$
(3) is
$$frac11-frac33+frac45-frac47+cdots=-3+frac13+4left(frac11-frac13+frac15-cdotsright),$$
(4) is
$$frac11-frac43+frac75-frac87+cdots=-7+frac43-frac15+8left(frac11-frac13+frac15-cdotsright),$$
etc.
In general, one gets
$$2^kfracpi4-sum_{j=0}^{k-1}(-1)^jfrac1{2j+1}
left(2^k-sum_{i=0}^j{kchoose i}right).$$
answered Dec 31 '18 at 16:38
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3057853%2fevaluate-sum-limits-n-0-infty-1n-sum-limits-j-0kk-choose-j-fra%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We will make use of the identity
$$
sum_{k=0}^n(-1)^kbinom{n}{k}frac1{k+alpha}=frac{n!,Gamma(alpha)}{Gamma(n+1+alpha)}tag1
$$
which can be proven by induction on $n$.
The sum in the question is
$$
begin{align}
&sum_{n=0}^infty(-1)^nsum_{j=0}^kbinom{k}{j}frac{(-1)^j}{2n+2j+1}tag2\
&=frac12sum_{n=0}^infty(-1)^nsum_{j=0}^kbinom{k}{j}frac{(-1)^j}{n+j+frac12}tag3\
&=frac12sum_{n=0}^infty(-1)^nfrac{k!,Gamma!left(n+frac12right)}{Gamma!left(n+k+frac32right)}tag4\
&=frac12sum_{n=0}^infty(-1)^nint_0^1(1-t)^kt^{n-frac12},mathrm{d}ttag5\
&=frac12int_0^1frac{(1-t)^k}{(1+t)sqrt{t}},mathrm{d}ttag6\
&=int_0^1frac{left(1-t^2right)^k}{1+t^2},mathrm{d}ttag7\
&=int_0^1frac{left(2-left(1+t^2right)right)^k}{1+t^2},mathrm{d}ttag8\
&=2^kleft(fracpi4+sum_{j=1}^kleft(-frac12right)^jbinom{k}{j}int_0^1left(1+t^2right)^{j-1},mathrm{d}tright)tag9\
&=2^kleft(fracpi4+sum_{j=1}^kleft(-frac12right)^jbinom{k}{j}sum_{i=0}^{j-1}binom{j-1}{i}frac1{2i+1}right)tag{10}\
&=2^kleft(fracpi4-frac12sum_{j=0}^{k-1}left(-frac12right)^jsum_{m=1}^kbinom{k-m}{j}sum_{i=0}^{j}binom{j}{i}frac1{2i+1}right)tag{11}\
&=2^kleft(fracpi4-frac12sum_{m=1}^ksum_{i=0}^{k-1}sum_{j=i}^{k-m}left(-frac12right)^jbinom{k-m}{i}binom{k-m-i}{j-i}frac1{2i+1}right)tag{12}\
&=2^kleft(fracpi4-frac12sum_{m=1}^ksum_{i=0}^{k-1}sum_{j=0}^{k-m-i}left(-frac12right)^{j+i}binom{k-m}{i}binom{k-m-i}{j}frac1{2i+1}right)tag{13}\
&=2^kleft(fracpi4-frac12sum_{m=1}^ksum_{i=0}^{k-1}left(-frac12right)^ibinom{k-m}{i}left(frac12right)^{k-m-i}frac1{2i+1}right)tag{14}\
&=2^kleft(fracpi4-frac14sum_{m=1}^ksum_{i=0}^{k-m}(-1)^ibinom{k-m}{i}left(frac12right)^{k-m}frac1{i+frac12}right)tag{15}\
&=2^kleft(fracpi4-frac14sum_{m=1}^kleft(frac12right)^{k-m}frac{(k-m)!,Gamma!left(frac12right)}{Gamma!left(k-m+frac32right)}right)tag{16}\
&=2^kleft(fracpi4-frac14sum_{m=0}^{k-1}left(frac12right)^mfrac{m!,Gamma!left(frac12right)}{Gamma!left(m+frac32right)}right)tag{17}\
&=2^kleft(fracpi4-frac12sum_{m=0}^{k-1}frac{m!}{(2m+1)!!}right)tag{18}\
end{align}
$$
Explanation:
$phantom{1}(3)$: bring $frac12$ out front
$phantom{1}(4)$: apply $(1)$
$phantom{1}(5)$: apply the Beta Function
$phantom{1}(6)$: sum in $n$
$phantom{1}(7)$: substitute $tmapsto t^2$
$phantom{1}(8)$: $1-t^2=2-left(1+t^2right)$
$phantom{1}(9)$: apply the Binomial Theorem
$(10)$: apply the Binomial Theorem and integrate
$(11)$: $binom{k}{j}=sumlimits_{m=1}^kbinom{k-m}{j-1}$ and substitute $jmapsto j+1$
$(12)$: reorder summation and $binom{k-m}{j}binom{j}{i}=binom{k-m}{i}binom{k-m-i}{j-i}$
$(13)$: substitute $jmapsto j+i$
$(14)$: apply the Binomial Theorem
$(15)$: shift factors of $2$
$(16)$: apply $(1)$
$(17)$: substitute $mmapsto k-m$
$(18)$: rewrite the Gamma functions as factorials
Thus,
$$
bbox[5px,border:2px solid #C0A000]{g(k)=-2^{k-1}sum_{m=0}^{k-1}frac{m!}{(2m+1)!!}}tag{19}
$$
$(4)$ says that asymptotically, the full sum from the question is $frac12sqrt{fracpi{k}}$, which vanishes as $ktoinfty$. Thus, the sum in $(18)$ must converge quickly to $fracpi2$. In fact, the sum in $(18)$ is the partial sum of the sum in this question whose limit is $fracpi2$.
answered Jan 1 at 14:22
robjohn♦robjohn
271k27317644
$endgroup$
add a comment |
$begingroup$
We will make use of the identity
$$
sum_{k=0}^n(-1)^kbinom{n}{k}frac1{k+alpha}=frac{n!,Gamma(alpha)}{Gamma(n+1+alpha)}tag1
$$
which can be proven by induction on $n$.
The sum in the question is
$$
begin{align}
&sum_{n=0}^infty(-1)^nsum_{j=0}^kbinom{k}{j}frac{(-1)^j}{2n+2j+1}tag2\
&=frac12sum_{n=0}^infty(-1)^nsum_{j=0}^kbinom{k}{j}frac{(-1)^j}{n+j+frac12}tag3\
&=frac12sum_{n=0}^infty(-1)^nfrac{k!,Gamma!left(n+frac12right)}{Gamma!left(n+k+frac32right)}tag4\
&=frac12sum_{n=0}^infty(-1)^nint_0^1(1-t)^kt^{n-frac12},mathrm{d}ttag5\
&=frac12int_0^1frac{(1-t)^k}{(1+t)sqrt{t}},mathrm{d}ttag6\
&=int_0^1frac{left(1-t^2right)^k}{1+t^2},mathrm{d}ttag7\
&=int_0^1frac{left(2-left(1+t^2right)right)^k}{1+t^2},mathrm{d}ttag8\
&=2^kleft(fracpi4+sum_{j=1}^kleft(-frac12right)^jbinom{k}{j}int_0^1left(1+t^2right)^{j-1},mathrm{d}tright)tag9\
&=2^kleft(fracpi4+sum_{j=1}^kleft(-frac12right)^jbinom{k}{j}sum_{i=0}^{j-1}binom{j-1}{i}frac1{2i+1}right)tag{10}\
&=2^kleft(fracpi4-frac12sum_{j=0}^{k-1}left(-frac12right)^jsum_{m=1}^kbinom{k-m}{j}sum_{i=0}^{j}binom{j}{i}frac1{2i+1}right)tag{11}\
&=2^kleft(fracpi4-frac12sum_{m=1}^ksum_{i=0}^{k-1}sum_{j=i}^{k-m}left(-frac12right)^jbinom{k-m}{i}binom{k-m-i}{j-i}frac1{2i+1}right)tag{12}\
&=2^kleft(fracpi4-frac12sum_{m=1}^ksum_{i=0}^{k-1}sum_{j=0}^{k-m-i}left(-frac12right)^{j+i}binom{k-m}{i}binom{k-m-i}{j}frac1{2i+1}right)tag{13}\
&=2^kleft(fracpi4-frac12sum_{m=1}^ksum_{i=0}^{k-1}left(-frac12right)^ibinom{k-m}{i}left(frac12right)^{k-m-i}frac1{2i+1}right)tag{14}\
&=2^kleft(fracpi4-frac14sum_{m=1}^ksum_{i=0}^{k-m}(-1)^ibinom{k-m}{i}left(frac12right)^{k-m}frac1{i+frac12}right)tag{15}\
&=2^kleft(fracpi4-frac14sum_{m=1}^kleft(frac12right)^{k-m}frac{(k-m)!,Gamma!left(frac12right)}{Gamma!left(k-m+frac32right)}right)tag{16}\
&=2^kleft(fracpi4-frac14sum_{m=0}^{k-1}left(frac12right)^mfrac{m!,Gamma!left(frac12right)}{Gamma!left(m+frac32right)}right)tag{17}\
&=2^kleft(fracpi4-frac12sum_{m=0}^{k-1}frac{m!}{(2m+1)!!}right)tag{18}\
end{align}
$$
Explanation:
$phantom{1}(3)$: bring $frac12$ out front
$phantom{1}(4)$: apply $(1)$
$phantom{1}(5)$: apply the Beta Function
$phantom{1}(6)$: sum in $n$
$phantom{1}(7)$: substitute $tmapsto t^2$
$phantom{1}(8)$: $1-t^2=2-left(1+t^2right)$
$phantom{1}(9)$: apply the Binomial Theorem
$(10)$: apply the Binomial Theorem and integrate
$(11)$: $binom{k}{j}=sumlimits_{m=1}^kbinom{k-m}{j-1}$ and substitute $jmapsto j+1$
$(12)$: reorder summation and $binom{k-m}{j}binom{j}{i}=binom{k-m}{i}binom{k-m-i}{j-i}$
$(13)$: substitute $jmapsto j+i$
$(14)$: apply the Binomial Theorem
$(15)$: shift factors of $2$
$(16)$: apply $(1)$
$(17)$: substitute $mmapsto k-m$
$(18)$: rewrite the Gamma functions as factorials
Thus,
$$
bbox[5px,border:2px solid #C0A000]{g(k)=-2^{k-1}sum_{m=0}^{k-1}frac{m!}{(2m+1)!!}}tag{19}
$$
$(4)$ says that asymptotically, the full sum from the question is $frac12sqrt{fracpi{k}}$, which vanishes as $ktoinfty$. Thus, the sum in $(18)$ must converge quickly to $fracpi2$. In fact, the sum in $(18)$ is the partial sum of the sum in this question whose limit is $fracpi2$.
answered Jan 1 at 14:22
robjohn♦robjohn
271k27317644
$endgroup$
add a comment |
$begingroup$
We will make use of the identity
$$
sum_{k=0}^n(-1)^kbinom{n}{k}frac1{k+alpha}=frac{n!,Gamma(alpha)}{Gamma(n+1+alpha)}tag1
$$
which can be proven by induction on $n$.
The sum in the question is
$$
begin{align}
&sum_{n=0}^infty(-1)^nsum_{j=0}^kbinom{k}{j}frac{(-1)^j}{2n+2j+1}tag2\
&=frac12sum_{n=0}^infty(-1)^nsum_{j=0}^kbinom{k}{j}frac{(-1)^j}{n+j+frac12}tag3\
&=frac12sum_{n=0}^infty(-1)^nfrac{k!,Gamma!left(n+frac12right)}{Gamma!left(n+k+frac32right)}tag4\
&=frac12sum_{n=0}^infty(-1)^nint_0^1(1-t)^kt^{n-frac12},mathrm{d}ttag5\
&=frac12int_0^1frac{(1-t)^k}{(1+t)sqrt{t}},mathrm{d}ttag6\
&=int_0^1frac{left(1-t^2right)^k}{1+t^2},mathrm{d}ttag7\
&=int_0^1frac{left(2-left(1+t^2right)right)^k}{1+t^2},mathrm{d}ttag8\
&=2^kleft(fracpi4+sum_{j=1}^kleft(-frac12right)^jbinom{k}{j}int_0^1left(1+t^2right)^{j-1},mathrm{d}tright)tag9\
&=2^kleft(fracpi4+sum_{j=1}^kleft(-frac12right)^jbinom{k}{j}sum_{i=0}^{j-1}binom{j-1}{i}frac1{2i+1}right)tag{10}\
&=2^kleft(fracpi4-frac12sum_{j=0}^{k-1}left(-frac12right)^jsum_{m=1}^kbinom{k-m}{j}sum_{i=0}^{j}binom{j}{i}frac1{2i+1}right)tag{11}\
&=2^kleft(fracpi4-frac12sum_{m=1}^ksum_{i=0}^{k-1}sum_{j=i}^{k-m}left(-frac12right)^jbinom{k-m}{i}binom{k-m-i}{j-i}frac1{2i+1}right)tag{12}\
&=2^kleft(fracpi4-frac12sum_{m=1}^ksum_{i=0}^{k-1}sum_{j=0}^{k-m-i}left(-frac12right)^{j+i}binom{k-m}{i}binom{k-m-i}{j}frac1{2i+1}right)tag{13}\
&=2^kleft(fracpi4-frac12sum_{m=1}^ksum_{i=0}^{k-1}left(-frac12right)^ibinom{k-m}{i}left(frac12right)^{k-m-i}frac1{2i+1}right)tag{14}\
&=2^kleft(fracpi4-frac14sum_{m=1}^ksum_{i=0}^{k-m}(-1)^ibinom{k-m}{i}left(frac12right)^{k-m}frac1{i+frac12}right)tag{15}\
&=2^kleft(fracpi4-frac14sum_{m=1}^kleft(frac12right)^{k-m}frac{(k-m)!,Gamma!left(frac12right)}{Gamma!left(k-m+frac32right)}right)tag{16}\
&=2^kleft(fracpi4-frac14sum_{m=0}^{k-1}left(frac12right)^mfrac{m!,Gamma!left(frac12right)}{Gamma!left(m+frac32right)}right)tag{17}\
&=2^kleft(fracpi4-frac12sum_{m=0}^{k-1}frac{m!}{(2m+1)!!}right)tag{18}\
end{align}
$$
Explanation:
$phantom{1}(3)$: bring $frac12$ out front
$phantom{1}(4)$: apply $(1)$
$phantom{1}(5)$: apply the Beta Function
$phantom{1}(6)$: sum in $n$
$phantom{1}(7)$: substitute $tmapsto t^2$
$phantom{1}(8)$: $1-t^2=2-left(1+t^2right)$
$phantom{1}(9)$: apply the Binomial Theorem
$(10)$: apply the Binomial Theorem and integrate
$(11)$: $binom{k}{j}=sumlimits_{m=1}^kbinom{k-m}{j-1}$ and substitute $jmapsto j+1$
$(12)$: reorder summation and $binom{k-m}{j}binom{j}{i}=binom{k-m}{i}binom{k-m-i}{j-i}$
$(13)$: substitute $jmapsto j+i$
$(14)$: apply the Binomial Theorem
$(15)$: shift factors of $2$
$(16)$: apply $(1)$
$(17)$: substitute $mmapsto k-m$
$(18)$: rewrite the Gamma functions as factorials
Thus,
$$
bbox[5px,border:2px solid #C0A000]{g(k)=-2^{k-1}sum_{m=0}^{k-1}frac{m!}{(2m+1)!!}}tag{19}
$$
$(4)$ says that asymptotically, the full sum from the question is $frac12sqrt{fracpi{k}}$, which vanishes as $ktoinfty$. Thus, the sum in $(18)$ must converge quickly to $fracpi2$. In fact, the sum in $(18)$ is the partial sum of the sum in this question whose limit is $fracpi2$.
answered Jan 1 at 14:22
robjohn♦robjohn
271k27317644
$endgroup$
We will make use of the identity
$$
sum_{k=0}^n(-1)^kbinom{n}{k}frac1{k+alpha}=frac{n!,Gamma(alpha)}{Gamma(n+1+alpha)}tag1
$$
which can be proven by induction on $n$.
The sum in the question is
$$
begin{align}
&sum_{n=0}^infty(-1)^nsum_{j=0}^kbinom{k}{j}frac{(-1)^j}{2n+2j+1}tag2\
&=frac12sum_{n=0}^infty(-1)^nsum_{j=0}^kbinom{k}{j}frac{(-1)^j}{n+j+frac12}tag3\
&=frac12sum_{n=0}^infty(-1)^nfrac{k!,Gamma!left(n+frac12right)}{Gamma!left(n+k+frac32right)}tag4\
&=frac12sum_{n=0}^infty(-1)^nint_0^1(1-t)^kt^{n-frac12},mathrm{d}ttag5\
&=frac12int_0^1frac{(1-t)^k}{(1+t)sqrt{t}},mathrm{d}ttag6\
&=int_0^1frac{left(1-t^2right)^k}{1+t^2},mathrm{d}ttag7\
&=int_0^1frac{left(2-left(1+t^2right)right)^k}{1+t^2},mathrm{d}ttag8\
&=2^kleft(fracpi4+sum_{j=1}^kleft(-frac12right)^jbinom{k}{j}int_0^1left(1+t^2right)^{j-1},mathrm{d}tright)tag9\
&=2^kleft(fracpi4+sum_{j=1}^kleft(-frac12right)^jbinom{k}{j}sum_{i=0}^{j-1}binom{j-1}{i}frac1{2i+1}right)tag{10}\
&=2^kleft(fracpi4-frac12sum_{j=0}^{k-1}left(-frac12right)^jsum_{m=1}^kbinom{k-m}{j}sum_{i=0}^{j}binom{j}{i}frac1{2i+1}right)tag{11}\
&=2^kleft(fracpi4-frac12sum_{m=1}^ksum_{i=0}^{k-1}sum_{j=i}^{k-m}left(-frac12right)^jbinom{k-m}{i}binom{k-m-i}{j-i}frac1{2i+1}right)tag{12}\
&=2^kleft(fracpi4-frac12sum_{m=1}^ksum_{i=0}^{k-1}sum_{j=0}^{k-m-i}left(-frac12right)^{j+i}binom{k-m}{i}binom{k-m-i}{j}frac1{2i+1}right)tag{13}\
&=2^kleft(fracpi4-frac12sum_{m=1}^ksum_{i=0}^{k-1}left(-frac12right)^ibinom{k-m}{i}left(frac12right)^{k-m-i}frac1{2i+1}right)tag{14}\
&=2^kleft(fracpi4-frac14sum_{m=1}^ksum_{i=0}^{k-m}(-1)^ibinom{k-m}{i}left(frac12right)^{k-m}frac1{i+frac12}right)tag{15}\
&=2^kleft(fracpi4-frac14sum_{m=1}^kleft(frac12right)^{k-m}frac{(k-m)!,Gamma!left(frac12right)}{Gamma!left(k-m+frac32right)}right)tag{16}\
&=2^kleft(fracpi4-frac14sum_{m=0}^{k-1}left(frac12right)^mfrac{m!,Gamma!left(frac12right)}{Gamma!left(m+frac32right)}right)tag{17}\
&=2^kleft(fracpi4-frac12sum_{m=0}^{k-1}frac{m!}{(2m+1)!!}right)tag{18}\
end{align}
$$
Explanation:
$phantom{1}(3)$: bring $frac12$ out front
$phantom{1}(4)$: apply $(1)$
$phantom{1}(5)$: apply the Beta Function
$phantom{1}(6)$: sum in $n$
$phantom{1}(7)$: substitute $tmapsto t^2$
$phantom{1}(8)$: $1-t^2=2-left(1+t^2right)$
$phantom{1}(9)$: apply the Binomial Theorem
$(10)$: apply the Binomial Theorem and integrate
$(11)$: $binom{k}{j}=sumlimits_{m=1}^kbinom{k-m}{j-1}$ and substitute $jmapsto j+1$
$(12)$: reorder summation and $binom{k-m}{j}binom{j}{i}=binom{k-m}{i}binom{k-m-i}{j-i}$
$(13)$: substitute $jmapsto j+i$
$(14)$: apply the Binomial Theorem
$(15)$: shift factors of $2$
$(16)$: apply $(1)$
$(17)$: substitute $mmapsto k-m$
$(18)$: rewrite the Gamma functions as factorials
Thus,
$$
bbox[5px,border:2px solid #C0A000]{g(k)=-2^{k-1}sum_{m=0}^{k-1}frac{m!}{(2m+1)!!}}tag{19}
$$
$(4)$ says that asymptotically, the full sum from the question is $frac12sqrt{fracpi{k}}$, which vanishes as $ktoinfty$. Thus, the sum in $(18)$ must converge quickly to $fracpi2$. In fact, the sum in $(18)$ is the partial sum of the sum in this question whose limit is $fracpi2$.
answered Jan 1 at 14:22
robjohn♦robjohn
271k27317644
edited Jan 2 at 2:47
answered Jan 1 at 14:22
robjohn♦robjohn
271k27317644
answered Jan 1 at 14:22
robjohn♦robjohn
271k27317644
answered Jan 1 at 14:22
robjohn♦robjohn
271k27317644
271k27317644
add a comment |
add a comment |
$begingroup$
Note that the partial fractions decomposition of the inverse
of the Rising Factorial gives:
$$
eqalign{
& left( {x - 1} right)^{,underline {, - left( {m + 1} right),} } = {1 over {x^{,overline {,m + 1,} } }}
= {{Gamma (x)} over {Gamma (x + m + 1)}} = ;quad left| {;0 le {rm integer }, m} right.quad = cr
& = sumlimits_{0, le ,k, le ,m} {{1 over {x + k}}left( {{{left( { - 1} right)^{,k} } over {left( {m - k} right)!k!}}} right)}
= sumlimits_{0, le ,k, le ,m} {{1 over {x + k}}left( {{{left( { - 1} right)^{,k} } over {m!}}binom{m}{k}
} right)} cr}
$$
So
$$
eqalign{
& sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
{1 over {2n + 2j + 1}}} = cr
& = {1 over 2}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
{1 over {left( {n + 1/2 + j} right)}}} = cr
& = {{k!} over {2left( {n + 1/2} right)^{,overline {,k + 1,} } }} cr}
$$
We have then that the sum in $n$ is
$$
eqalign{
& S(k) = sumlimits_{0, le ,n} {left( { - 1} right)^{,n} sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right){1 over {2n + 2j + 1}}} } = cr
& = {1 over 2}sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{k!} over {left( {n + 1/2} right)^{,overline {,k + 1,} } }}} = cr
& = {1 over {2left( {k + 1} right)}}sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{1^{,overline {,k + 1,} } }
over {left( {1 + n - 1/2} right)^{,overline {,k + 1,} } }}} = cr
& = {1 over {2left( {k + 1} right)}}sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{1^{,overline {,n - 1/2,} } }
over {left( {2 + k} right)^{,overline {,n - 1/2,} } }}} = cr
& = {{1^{,overline {, - 1/2,} } } over {2left( {k + 1} right)left( {2 + k} right)^{,overline {, - 1/2,} } }}
sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{left( {1/2} right)^{,overline {,n,} } } over {left( {3/2 + k} right)^{,overline {,n,} } }}} = cr
& = {{Gamma left( {1/2} right)Gamma left( {2 + k} right)} over {2left( {k + 1} right)Gamma left( 1 right)Gamma left( {3/2 + k} right) }}
sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{left( {1/2} right)^{,overline {,n,} } } over {left( {3/2 + k} right)^{,overline {,n,} } }}} = cr
& = {{Gamma left( {1/2} right)Gamma left( {1 + k} right)} over {2,Gamma left( {3/2 + k} right)}}{}_2F_{,1} left( {left. {matrix{
{1/2;,;1} cr
{k + 3/2} cr
} ;} right| - 1} right) = cr
& = 2^{,2,k} {{left( {k!} right)^{,2} } over {left( {2k + 1} right)!}};;{}_2F_{,1} left( {left. {matrix{
{1/2;,;1} cr
{k + 3/2} cr
} ;} right| - 1} right) cr}
$$
where in the mid of the derivation we have used the identity
$$
{{z^{,overline {,w,} } } over {left( {z + a} right)^{,overline {,w,} } }} = {{z^{,overline {,a,} } } over {left( {z + w} right)^{,overline {,a,} } }}
$$
and at the end the duplication formula for Gamma
$$
Gamma left( {2,z} right) = {{2^{,2,z - 1} } over {sqrt pi }}Gamma left( z right)Gamma left( {z + 1/2} right) = 2^{,2,z - 1} {{Gamma left( z right)Gamma left( {z + 1/2} right)} over {Gamma left( {1/2} right)}}
$$
Alternative Approach
Actually there is a more straight alternative aproach through the Digamma Function
$$
eqalign{
& S(k) = sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {1 over {2n + 2j + 1}}} } = cr
& = sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)sumlimits_{0, le ,m} {left( {{1 over {4m + 2j + 1}} - {1 over {4m + 2j + 3}}} right)} } = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)sumlimits_{0, le ,m} {left( {{1 over {m + j/2 + 1/4}} - {1 over {m + j/2 + 3/4}}} right)} } = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)left( {psi left( {j/2 + 3/4} right) - psi left( {j/2 + 1/4} right)} right)} = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,} {left( matrix{
k cr
2j cr} right)left( {psi left( {j + 3/4} right) - psi left( {j + 1/4} right)} right) - left( matrix{
k cr
2j + 1 cr} right)left( {psi left( {j + 5/4} right) - psi left( {j + 3/4} right)} right)} = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,} {left( matrix{
k + 1 cr
2j + 1 cr} right)psi left( {j + 3/4} right) - left( matrix{
k cr
2j cr} right)psi left( {j + 1/4} right) - left( matrix{
k cr
2j + 1 cr} right)psi left( {j + 1 + 1/4} right)} = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,} {left( matrix{
k + 1 cr
2j + 1 cr} right)left( {psi left( {j + 3/4} right) - psi left( {j + 1/4} right)} right) - left( matrix{
k cr
2j + 1 cr} right){1 over {j + 1/4}}} cr}
$$
Since we can write (re. for instance to this page)
$$
eqalign{
& psi left( {j + 3/4} right) = 4sumlimits_{0, le ,l, le ,j - 1,} {{1 over {4,l + 3}}} + {pi over 2} - ln 8 - gamma
= psi left( {j + 3/4} right) - left( {psi left( {3/4} right) - left( {{pi over 2} - ln 8 - gamma } right)} right) cr
& psi left( {j + 1/4} right) = 4sumlimits_{0, le ,l, le ,j - 1,} {{1 over {4,l + 1}}} - {pi over 2} - ln 8 - gamma
= psi left( {j + 1/4} right) - left( {psi left( {1/4} right) - left( { - {pi over 2} - ln 8 - gamma } right)} right) cr}
$$
we arrive to express the sum as
$$
eqalign{
& S(k) = sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {1 over {2n + 2j + 1}}} } = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
left( {psi left( {j/2 + 3/4} right) - psi left( {j/2 + 1/4} right)} right)} = cr
& = {pi over 4}2^{,k} - 2sumlimits_{left( {0, le } right),j,left( { le ,k} right),} {binom{k+1}{2j+1}
left( {sumlimits_{0, le ,l, le ,j - 1,} {{1 over {left( {4,l + 3} right)left( {4,l + 1} right)}}} } right)}
- sumlimits_{left( {0, le } right),j,left( { le ,left( {k - 1} right)/2} right),} {binom{k}{2j+1}{1 over {4j + 1}}} = cr
& = {pi over 4}2^{,k} - 2sumlimits_{left( {0, le } right),j,left( { le ,k} right),} {binom{k+1}{2j+1}
left( {sumlimits_{0, le ,l, le ,j - 1,} {{1 over {left( {4,l + 2} right)^{,2} - 1}}} } right)}
- sumlimits_{left( {0, le } right),j,left( { le ,left( {k - 1} right)/2} right),} {binom{k}{2j+1}{1 over {4j + 1}}} cr}
$$
Interesting the comparison of this expression with the one given by
@robjohn
answered Jan 1 at 17:31
G CabG Cab
20.5k31342
$endgroup$
add a comment |
$begingroup$
Note that the partial fractions decomposition of the inverse
of the Rising Factorial gives:
$$
eqalign{
& left( {x - 1} right)^{,underline {, - left( {m + 1} right),} } = {1 over {x^{,overline {,m + 1,} } }}
= {{Gamma (x)} over {Gamma (x + m + 1)}} = ;quad left| {;0 le {rm integer }, m} right.quad = cr
& = sumlimits_{0, le ,k, le ,m} {{1 over {x + k}}left( {{{left( { - 1} right)^{,k} } over {left( {m - k} right)!k!}}} right)}
= sumlimits_{0, le ,k, le ,m} {{1 over {x + k}}left( {{{left( { - 1} right)^{,k} } over {m!}}binom{m}{k}
} right)} cr}
$$
So
$$
eqalign{
& sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
{1 over {2n + 2j + 1}}} = cr
& = {1 over 2}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
{1 over {left( {n + 1/2 + j} right)}}} = cr
& = {{k!} over {2left( {n + 1/2} right)^{,overline {,k + 1,} } }} cr}
$$
We have then that the sum in $n$ is
$$
eqalign{
& S(k) = sumlimits_{0, le ,n} {left( { - 1} right)^{,n} sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right){1 over {2n + 2j + 1}}} } = cr
& = {1 over 2}sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{k!} over {left( {n + 1/2} right)^{,overline {,k + 1,} } }}} = cr
& = {1 over {2left( {k + 1} right)}}sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{1^{,overline {,k + 1,} } }
over {left( {1 + n - 1/2} right)^{,overline {,k + 1,} } }}} = cr
& = {1 over {2left( {k + 1} right)}}sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{1^{,overline {,n - 1/2,} } }
over {left( {2 + k} right)^{,overline {,n - 1/2,} } }}} = cr
& = {{1^{,overline {, - 1/2,} } } over {2left( {k + 1} right)left( {2 + k} right)^{,overline {, - 1/2,} } }}
sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{left( {1/2} right)^{,overline {,n,} } } over {left( {3/2 + k} right)^{,overline {,n,} } }}} = cr
& = {{Gamma left( {1/2} right)Gamma left( {2 + k} right)} over {2left( {k + 1} right)Gamma left( 1 right)Gamma left( {3/2 + k} right) }}
sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{left( {1/2} right)^{,overline {,n,} } } over {left( {3/2 + k} right)^{,overline {,n,} } }}} = cr
& = {{Gamma left( {1/2} right)Gamma left( {1 + k} right)} over {2,Gamma left( {3/2 + k} right)}}{}_2F_{,1} left( {left. {matrix{
{1/2;,;1} cr
{k + 3/2} cr
} ;} right| - 1} right) = cr
& = 2^{,2,k} {{left( {k!} right)^{,2} } over {left( {2k + 1} right)!}};;{}_2F_{,1} left( {left. {matrix{
{1/2;,;1} cr
{k + 3/2} cr
} ;} right| - 1} right) cr}
$$
where in the mid of the derivation we have used the identity
$$
{{z^{,overline {,w,} } } over {left( {z + a} right)^{,overline {,w,} } }} = {{z^{,overline {,a,} } } over {left( {z + w} right)^{,overline {,a,} } }}
$$
and at the end the duplication formula for Gamma
$$
Gamma left( {2,z} right) = {{2^{,2,z - 1} } over {sqrt pi }}Gamma left( z right)Gamma left( {z + 1/2} right) = 2^{,2,z - 1} {{Gamma left( z right)Gamma left( {z + 1/2} right)} over {Gamma left( {1/2} right)}}
$$
Alternative Approach
Actually there is a more straight alternative aproach through the Digamma Function
$$
eqalign{
& S(k) = sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {1 over {2n + 2j + 1}}} } = cr
& = sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)sumlimits_{0, le ,m} {left( {{1 over {4m + 2j + 1}} - {1 over {4m + 2j + 3}}} right)} } = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)sumlimits_{0, le ,m} {left( {{1 over {m + j/2 + 1/4}} - {1 over {m + j/2 + 3/4}}} right)} } = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)left( {psi left( {j/2 + 3/4} right) - psi left( {j/2 + 1/4} right)} right)} = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,} {left( matrix{
k cr
2j cr} right)left( {psi left( {j + 3/4} right) - psi left( {j + 1/4} right)} right) - left( matrix{
k cr
2j + 1 cr} right)left( {psi left( {j + 5/4} right) - psi left( {j + 3/4} right)} right)} = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,} {left( matrix{
k + 1 cr
2j + 1 cr} right)psi left( {j + 3/4} right) - left( matrix{
k cr
2j cr} right)psi left( {j + 1/4} right) - left( matrix{
k cr
2j + 1 cr} right)psi left( {j + 1 + 1/4} right)} = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,} {left( matrix{
k + 1 cr
2j + 1 cr} right)left( {psi left( {j + 3/4} right) - psi left( {j + 1/4} right)} right) - left( matrix{
k cr
2j + 1 cr} right){1 over {j + 1/4}}} cr}
$$
Since we can write (re. for instance to this page)
$$
eqalign{
& psi left( {j + 3/4} right) = 4sumlimits_{0, le ,l, le ,j - 1,} {{1 over {4,l + 3}}} + {pi over 2} - ln 8 - gamma
= psi left( {j + 3/4} right) - left( {psi left( {3/4} right) - left( {{pi over 2} - ln 8 - gamma } right)} right) cr
& psi left( {j + 1/4} right) = 4sumlimits_{0, le ,l, le ,j - 1,} {{1 over {4,l + 1}}} - {pi over 2} - ln 8 - gamma
= psi left( {j + 1/4} right) - left( {psi left( {1/4} right) - left( { - {pi over 2} - ln 8 - gamma } right)} right) cr}
$$
we arrive to express the sum as
$$
eqalign{
& S(k) = sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {1 over {2n + 2j + 1}}} } = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
left( {psi left( {j/2 + 3/4} right) - psi left( {j/2 + 1/4} right)} right)} = cr
& = {pi over 4}2^{,k} - 2sumlimits_{left( {0, le } right),j,left( { le ,k} right),} {binom{k+1}{2j+1}
left( {sumlimits_{0, le ,l, le ,j - 1,} {{1 over {left( {4,l + 3} right)left( {4,l + 1} right)}}} } right)}
- sumlimits_{left( {0, le } right),j,left( { le ,left( {k - 1} right)/2} right),} {binom{k}{2j+1}{1 over {4j + 1}}} = cr
& = {pi over 4}2^{,k} - 2sumlimits_{left( {0, le } right),j,left( { le ,k} right),} {binom{k+1}{2j+1}
left( {sumlimits_{0, le ,l, le ,j - 1,} {{1 over {left( {4,l + 2} right)^{,2} - 1}}} } right)}
- sumlimits_{left( {0, le } right),j,left( { le ,left( {k - 1} right)/2} right),} {binom{k}{2j+1}{1 over {4j + 1}}} cr}
$$
Interesting the comparison of this expression with the one given by
@robjohn
answered Jan 1 at 17:31
G CabG Cab
20.5k31342
$endgroup$
add a comment |
$begingroup$
Note that the partial fractions decomposition of the inverse
of the Rising Factorial gives:
$$
eqalign{
& left( {x - 1} right)^{,underline {, - left( {m + 1} right),} } = {1 over {x^{,overline {,m + 1,} } }}
= {{Gamma (x)} over {Gamma (x + m + 1)}} = ;quad left| {;0 le {rm integer }, m} right.quad = cr
& = sumlimits_{0, le ,k, le ,m} {{1 over {x + k}}left( {{{left( { - 1} right)^{,k} } over {left( {m - k} right)!k!}}} right)}
= sumlimits_{0, le ,k, le ,m} {{1 over {x + k}}left( {{{left( { - 1} right)^{,k} } over {m!}}binom{m}{k}
} right)} cr}
$$
So
$$
eqalign{
& sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
{1 over {2n + 2j + 1}}} = cr
& = {1 over 2}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
{1 over {left( {n + 1/2 + j} right)}}} = cr
& = {{k!} over {2left( {n + 1/2} right)^{,overline {,k + 1,} } }} cr}
$$
We have then that the sum in $n$ is
$$
eqalign{
& S(k) = sumlimits_{0, le ,n} {left( { - 1} right)^{,n} sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right){1 over {2n + 2j + 1}}} } = cr
& = {1 over 2}sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{k!} over {left( {n + 1/2} right)^{,overline {,k + 1,} } }}} = cr
& = {1 over {2left( {k + 1} right)}}sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{1^{,overline {,k + 1,} } }
over {left( {1 + n - 1/2} right)^{,overline {,k + 1,} } }}} = cr
& = {1 over {2left( {k + 1} right)}}sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{1^{,overline {,n - 1/2,} } }
over {left( {2 + k} right)^{,overline {,n - 1/2,} } }}} = cr
& = {{1^{,overline {, - 1/2,} } } over {2left( {k + 1} right)left( {2 + k} right)^{,overline {, - 1/2,} } }}
sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{left( {1/2} right)^{,overline {,n,} } } over {left( {3/2 + k} right)^{,overline {,n,} } }}} = cr
& = {{Gamma left( {1/2} right)Gamma left( {2 + k} right)} over {2left( {k + 1} right)Gamma left( 1 right)Gamma left( {3/2 + k} right) }}
sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{left( {1/2} right)^{,overline {,n,} } } over {left( {3/2 + k} right)^{,overline {,n,} } }}} = cr
& = {{Gamma left( {1/2} right)Gamma left( {1 + k} right)} over {2,Gamma left( {3/2 + k} right)}}{}_2F_{,1} left( {left. {matrix{
{1/2;,;1} cr
{k + 3/2} cr
} ;} right| - 1} right) = cr
& = 2^{,2,k} {{left( {k!} right)^{,2} } over {left( {2k + 1} right)!}};;{}_2F_{,1} left( {left. {matrix{
{1/2;,;1} cr
{k + 3/2} cr
} ;} right| - 1} right) cr}
$$
where in the mid of the derivation we have used the identity
$$
{{z^{,overline {,w,} } } over {left( {z + a} right)^{,overline {,w,} } }} = {{z^{,overline {,a,} } } over {left( {z + w} right)^{,overline {,a,} } }}
$$
and at the end the duplication formula for Gamma
$$
Gamma left( {2,z} right) = {{2^{,2,z - 1} } over {sqrt pi }}Gamma left( z right)Gamma left( {z + 1/2} right) = 2^{,2,z - 1} {{Gamma left( z right)Gamma left( {z + 1/2} right)} over {Gamma left( {1/2} right)}}
$$
Alternative Approach
Actually there is a more straight alternative aproach through the Digamma Function
$$
eqalign{
& S(k) = sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {1 over {2n + 2j + 1}}} } = cr
& = sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)sumlimits_{0, le ,m} {left( {{1 over {4m + 2j + 1}} - {1 over {4m + 2j + 3}}} right)} } = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)sumlimits_{0, le ,m} {left( {{1 over {m + j/2 + 1/4}} - {1 over {m + j/2 + 3/4}}} right)} } = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)left( {psi left( {j/2 + 3/4} right) - psi left( {j/2 + 1/4} right)} right)} = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,} {left( matrix{
k cr
2j cr} right)left( {psi left( {j + 3/4} right) - psi left( {j + 1/4} right)} right) - left( matrix{
k cr
2j + 1 cr} right)left( {psi left( {j + 5/4} right) - psi left( {j + 3/4} right)} right)} = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,} {left( matrix{
k + 1 cr
2j + 1 cr} right)psi left( {j + 3/4} right) - left( matrix{
k cr
2j cr} right)psi left( {j + 1/4} right) - left( matrix{
k cr
2j + 1 cr} right)psi left( {j + 1 + 1/4} right)} = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,} {left( matrix{
k + 1 cr
2j + 1 cr} right)left( {psi left( {j + 3/4} right) - psi left( {j + 1/4} right)} right) - left( matrix{
k cr
2j + 1 cr} right){1 over {j + 1/4}}} cr}
$$
Since we can write (re. for instance to this page)
$$
eqalign{
& psi left( {j + 3/4} right) = 4sumlimits_{0, le ,l, le ,j - 1,} {{1 over {4,l + 3}}} + {pi over 2} - ln 8 - gamma
= psi left( {j + 3/4} right) - left( {psi left( {3/4} right) - left( {{pi over 2} - ln 8 - gamma } right)} right) cr
& psi left( {j + 1/4} right) = 4sumlimits_{0, le ,l, le ,j - 1,} {{1 over {4,l + 1}}} - {pi over 2} - ln 8 - gamma
= psi left( {j + 1/4} right) - left( {psi left( {1/4} right) - left( { - {pi over 2} - ln 8 - gamma } right)} right) cr}
$$
we arrive to express the sum as
$$
eqalign{
& S(k) = sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {1 over {2n + 2j + 1}}} } = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
left( {psi left( {j/2 + 3/4} right) - psi left( {j/2 + 1/4} right)} right)} = cr
& = {pi over 4}2^{,k} - 2sumlimits_{left( {0, le } right),j,left( { le ,k} right),} {binom{k+1}{2j+1}
left( {sumlimits_{0, le ,l, le ,j - 1,} {{1 over {left( {4,l + 3} right)left( {4,l + 1} right)}}} } right)}
- sumlimits_{left( {0, le } right),j,left( { le ,left( {k - 1} right)/2} right),} {binom{k}{2j+1}{1 over {4j + 1}}} = cr
& = {pi over 4}2^{,k} - 2sumlimits_{left( {0, le } right),j,left( { le ,k} right),} {binom{k+1}{2j+1}
left( {sumlimits_{0, le ,l, le ,j - 1,} {{1 over {left( {4,l + 2} right)^{,2} - 1}}} } right)}
- sumlimits_{left( {0, le } right),j,left( { le ,left( {k - 1} right)/2} right),} {binom{k}{2j+1}{1 over {4j + 1}}} cr}
$$
Interesting the comparison of this expression with the one given by
@robjohn
answered Jan 1 at 17:31
G CabG Cab
20.5k31342
$endgroup$
Note that the partial fractions decomposition of the inverse
of the Rising Factorial gives:
$$
eqalign{
& left( {x - 1} right)^{,underline {, - left( {m + 1} right),} } = {1 over {x^{,overline {,m + 1,} } }}
= {{Gamma (x)} over {Gamma (x + m + 1)}} = ;quad left| {;0 le {rm integer }, m} right.quad = cr
& = sumlimits_{0, le ,k, le ,m} {{1 over {x + k}}left( {{{left( { - 1} right)^{,k} } over {left( {m - k} right)!k!}}} right)}
= sumlimits_{0, le ,k, le ,m} {{1 over {x + k}}left( {{{left( { - 1} right)^{,k} } over {m!}}binom{m}{k}
} right)} cr}
$$
So
$$
eqalign{
& sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
{1 over {2n + 2j + 1}}} = cr
& = {1 over 2}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
{1 over {left( {n + 1/2 + j} right)}}} = cr
& = {{k!} over {2left( {n + 1/2} right)^{,overline {,k + 1,} } }} cr}
$$
We have then that the sum in $n$ is
$$
eqalign{
& S(k) = sumlimits_{0, le ,n} {left( { - 1} right)^{,n} sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right){1 over {2n + 2j + 1}}} } = cr
& = {1 over 2}sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{k!} over {left( {n + 1/2} right)^{,overline {,k + 1,} } }}} = cr
& = {1 over {2left( {k + 1} right)}}sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{1^{,overline {,k + 1,} } }
over {left( {1 + n - 1/2} right)^{,overline {,k + 1,} } }}} = cr
& = {1 over {2left( {k + 1} right)}}sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{1^{,overline {,n - 1/2,} } }
over {left( {2 + k} right)^{,overline {,n - 1/2,} } }}} = cr
& = {{1^{,overline {, - 1/2,} } } over {2left( {k + 1} right)left( {2 + k} right)^{,overline {, - 1/2,} } }}
sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{left( {1/2} right)^{,overline {,n,} } } over {left( {3/2 + k} right)^{,overline {,n,} } }}} = cr
& = {{Gamma left( {1/2} right)Gamma left( {2 + k} right)} over {2left( {k + 1} right)Gamma left( 1 right)Gamma left( {3/2 + k} right) }}
sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{left( {1/2} right)^{,overline {,n,} } } over {left( {3/2 + k} right)^{,overline {,n,} } }}} = cr
& = {{Gamma left( {1/2} right)Gamma left( {1 + k} right)} over {2,Gamma left( {3/2 + k} right)}}{}_2F_{,1} left( {left. {matrix{
{1/2;,;1} cr
{k + 3/2} cr
} ;} right| - 1} right) = cr
& = 2^{,2,k} {{left( {k!} right)^{,2} } over {left( {2k + 1} right)!}};;{}_2F_{,1} left( {left. {matrix{
{1/2;,;1} cr
{k + 3/2} cr
} ;} right| - 1} right) cr}
$$
where in the mid of the derivation we have used the identity
$$
{{z^{,overline {,w,} } } over {left( {z + a} right)^{,overline {,w,} } }} = {{z^{,overline {,a,} } } over {left( {z + w} right)^{,overline {,a,} } }}
$$
and at the end the duplication formula for Gamma
$$
Gamma left( {2,z} right) = {{2^{,2,z - 1} } over {sqrt pi }}Gamma left( z right)Gamma left( {z + 1/2} right) = 2^{,2,z - 1} {{Gamma left( z right)Gamma left( {z + 1/2} right)} over {Gamma left( {1/2} right)}}
$$
Alternative Approach
Actually there is a more straight alternative aproach through the Digamma Function
$$
eqalign{
& S(k) = sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {1 over {2n + 2j + 1}}} } = cr
& = sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)sumlimits_{0, le ,m} {left( {{1 over {4m + 2j + 1}} - {1 over {4m + 2j + 3}}} right)} } = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)sumlimits_{0, le ,m} {left( {{1 over {m + j/2 + 1/4}} - {1 over {m + j/2 + 3/4}}} right)} } = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)left( {psi left( {j/2 + 3/4} right) - psi left( {j/2 + 1/4} right)} right)} = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,} {left( matrix{
k cr
2j cr} right)left( {psi left( {j + 3/4} right) - psi left( {j + 1/4} right)} right) - left( matrix{
k cr
2j + 1 cr} right)left( {psi left( {j + 5/4} right) - psi left( {j + 3/4} right)} right)} = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,} {left( matrix{
k + 1 cr
2j + 1 cr} right)psi left( {j + 3/4} right) - left( matrix{
k cr
2j cr} right)psi left( {j + 1/4} right) - left( matrix{
k cr
2j + 1 cr} right)psi left( {j + 1 + 1/4} right)} = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,} {left( matrix{
k + 1 cr
2j + 1 cr} right)left( {psi left( {j + 3/4} right) - psi left( {j + 1/4} right)} right) - left( matrix{
k cr
2j + 1 cr} right){1 over {j + 1/4}}} cr}
$$
Since we can write (re. for instance to this page)
$$
eqalign{
& psi left( {j + 3/4} right) = 4sumlimits_{0, le ,l, le ,j - 1,} {{1 over {4,l + 3}}} + {pi over 2} - ln 8 - gamma
= psi left( {j + 3/4} right) - left( {psi left( {3/4} right) - left( {{pi over 2} - ln 8 - gamma } right)} right) cr
& psi left( {j + 1/4} right) = 4sumlimits_{0, le ,l, le ,j - 1,} {{1 over {4,l + 1}}} - {pi over 2} - ln 8 - gamma
= psi left( {j + 1/4} right) - left( {psi left( {1/4} right) - left( { - {pi over 2} - ln 8 - gamma } right)} right) cr}
$$
we arrive to express the sum as
$$
eqalign{
& S(k) = sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {1 over {2n + 2j + 1}}} } = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
left( {psi left( {j/2 + 3/4} right) - psi left( {j/2 + 1/4} right)} right)} = cr
& = {pi over 4}2^{,k} - 2sumlimits_{left( {0, le } right),j,left( { le ,k} right),} {binom{k+1}{2j+1}
left( {sumlimits_{0, le ,l, le ,j - 1,} {{1 over {left( {4,l + 3} right)left( {4,l + 1} right)}}} } right)}
- sumlimits_{left( {0, le } right),j,left( { le ,left( {k - 1} right)/2} right),} {binom{k}{2j+1}{1 over {4j + 1}}} = cr
& = {pi over 4}2^{,k} - 2sumlimits_{left( {0, le } right),j,left( { le ,k} right),} {binom{k+1}{2j+1}
left( {sumlimits_{0, le ,l, le ,j - 1,} {{1 over {left( {4,l + 2} right)^{,2} - 1}}} } right)}
- sumlimits_{left( {0, le } right),j,left( { le ,left( {k - 1} right)/2} right),} {binom{k}{2j+1}{1 over {4j + 1}}} cr}
$$
Interesting the comparison of this expression with the one given by
@robjohn
answered Jan 1 at 17:31
G CabG Cab
20.5k31342
edited Jan 3 at 11:57
answered Jan 1 at 17:31
G CabG Cab
20.5k31342
answered Jan 1 at 17:31
G CabG Cab
20.5k31342
answered Jan 1 at 17:31
G CabG Cab
20.5k31342
20.5k31342
add a comment |
add a comment |
$begingroup$
(2) is
$$frac11-frac23+frac25-frac27+cdots=-1+2left(frac11-frac13+frac15-cdotsright),$$
(3) is
$$frac11-frac33+frac45-frac47+cdots=-3+frac13+4left(frac11-frac13+frac15-cdotsright),$$
(4) is
$$frac11-frac43+frac75-frac87+cdots=-7+frac43-frac15+8left(frac11-frac13+frac15-cdotsright),$$
etc.
In general, one gets
$$2^kfracpi4-sum_{j=0}^{k-1}(-1)^jfrac1{2j+1}
left(2^k-sum_{i=0}^j{kchoose i}right).$$
answered Dec 31 '18 at 16:38
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
add a comment |
$begingroup$
(2) is
$$frac11-frac23+frac25-frac27+cdots=-1+2left(frac11-frac13+frac15-cdotsright),$$
(3) is
$$frac11-frac33+frac45-frac47+cdots=-3+frac13+4left(frac11-frac13+frac15-cdotsright),$$
(4) is
$$frac11-frac43+frac75-frac87+cdots=-7+frac43-frac15+8left(frac11-frac13+frac15-cdotsright),$$
etc.
In general, one gets
$$2^kfracpi4-sum_{j=0}^{k-1}(-1)^jfrac1{2j+1}
left(2^k-sum_{i=0}^j{kchoose i}right).$$
answered Dec 31 '18 at 16:38
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
add a comment |
$begingroup$
(2) is
$$frac11-frac23+frac25-frac27+cdots=-1+2left(frac11-frac13+frac15-cdotsright),$$
(3) is
$$frac11-frac33+frac45-frac47+cdots=-3+frac13+4left(frac11-frac13+frac15-cdotsright),$$
(4) is
$$frac11-frac43+frac75-frac87+cdots=-7+frac43-frac15+8left(frac11-frac13+frac15-cdotsright),$$
etc.
In general, one gets
$$2^kfracpi4-sum_{j=0}^{k-1}(-1)^jfrac1{2j+1}
left(2^k-sum_{i=0}^j{kchoose i}right).$$
answered Dec 31 '18 at 16:38
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
(2) is
$$frac11-frac23+frac25-frac27+cdots=-1+2left(frac11-frac13+frac15-cdotsright),$$
(3) is
$$frac11-frac33+frac45-frac47+cdots=-3+frac13+4left(frac11-frac13+frac15-cdotsright),$$
(4) is
$$frac11-frac43+frac75-frac87+cdots=-7+frac43-frac15+8left(frac11-frac13+frac15-cdotsright),$$
etc.
In general, one gets
$$2^kfracpi4-sum_{j=0}^{k-1}(-1)^jfrac1{2j+1}
left(2^k-sum_{i=0}^j{kchoose i}right).$$
answered Dec 31 '18 at 16:38
Lord Shark the UnknownLord Shark the Unknown
109k1163136
edited Dec 31 '18 at 16:44
answered Dec 31 '18 at 16:38
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered Dec 31 '18 at 16:38
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered Dec 31 '18 at 16:38
Lord Shark the UnknownLord Shark the Unknown
109k1163136
109k1163136
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3057853%2fevaluate-sum-limits-n-0-infty-1n-sum-limits-j-0kk-choose-j-fra%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
