Evaluating the Cauchy product of $sum_{n=0}^{infty}frac{(-1)^{n+1}}{n+1}$ and $sum_{n=0}^{infty}frac{1}{3^n}...
$begingroup$
Using the Mertens' theorem for Cauchy products we know that the Cauchy product of series
$$
sum_{n=0}^{infty}frac{(-1)^{n+1}}{n+1}qquadtext{and}qquadsum_{n=0}^{infty}frac{1}{3^n}
$$ does converge. But how can we try to find the value of this sum?
What I only know is that we can write that this sum equals to
$$
S=sum_{n=0}^{infty}c_nqquadtext{where}qquad c_n=sum_{k=0}^{n}frac{1}{3^{n-k}}frac{(-1)^{k+1}}{k+1}
$$
but I'm not able to find any way to calculate the sum. Any hints?
sequences-and-series summation
edited Dec 23 '18 at 12:30
Blue
49.3k870157
asked Dec 23 '18 at 12:23
avan1235avan1235
3578
$endgroup$
add a comment |
$begingroup$
Using the Mertens' theorem for Cauchy products we know that the Cauchy product of series
$$
sum_{n=0}^{infty}frac{(-1)^{n+1}}{n+1}qquadtext{and}qquadsum_{n=0}^{infty}frac{1}{3^n}
$$ does converge. But how can we try to find the value of this sum?
What I only know is that we can write that this sum equals to
$$
S=sum_{n=0}^{infty}c_nqquadtext{where}qquad c_n=sum_{k=0}^{n}frac{1}{3^{n-k}}frac{(-1)^{k+1}}{k+1}
$$
but I'm not able to find any way to calculate the sum. Any hints?
sequences-and-series summation
edited Dec 23 '18 at 12:30
Blue
49.3k870157
asked Dec 23 '18 at 12:23
avan1235avan1235
3578
$endgroup$
5
$begingroup$
If I understand correctly, by MERTENS' Theorem, $S$ is equal to the product of the of the initial sums, right? The first is $-ln(2)$ and the second is $frac{3}{2}$ by the geometric series.
$endgroup$
– Viktor Glombik
Dec 23 '18 at 12:38
1
$begingroup$
Yes but how to find it using only the definition of Cauchy product of series?
$endgroup$
– avan1235
Dec 23 '18 at 12:44
$begingroup$
If you were presented with this sum as a $S$ and asked to evaluate it. I would think that decomposing it (via cauchy product) would be the easiest way of coming to a conclusion.
$endgroup$
– Mason
Dec 23 '18 at 16:11
add a comment |
$begingroup$
Using the Mertens' theorem for Cauchy products we know that the Cauchy product of series
$$
sum_{n=0}^{infty}frac{(-1)^{n+1}}{n+1}qquadtext{and}qquadsum_{n=0}^{infty}frac{1}{3^n}
$$ does converge. But how can we try to find the value of this sum?
What I only know is that we can write that this sum equals to
$$
S=sum_{n=0}^{infty}c_nqquadtext{where}qquad c_n=sum_{k=0}^{n}frac{1}{3^{n-k}}frac{(-1)^{k+1}}{k+1}
$$
but I'm not able to find any way to calculate the sum. Any hints?
sequences-and-series summation
edited Dec 23 '18 at 12:30
Blue
49.3k870157
asked Dec 23 '18 at 12:23
avan1235avan1235
3578
$endgroup$
Using the Mertens' theorem for Cauchy products we know that the Cauchy product of series
$$
sum_{n=0}^{infty}frac{(-1)^{n+1}}{n+1}qquadtext{and}qquadsum_{n=0}^{infty}frac{1}{3^n}
$$ does converge. But how can we try to find the value of this sum?
What I only know is that we can write that this sum equals to
$$
S=sum_{n=0}^{infty}c_nqquadtext{where}qquad c_n=sum_{k=0}^{n}frac{1}{3^{n-k}}frac{(-1)^{k+1}}{k+1}
$$
but I'm not able to find any way to calculate the sum. Any hints?
sequences-and-series summation
sequences-and-series summation
edited Dec 23 '18 at 12:30
Blue
49.3k870157
asked Dec 23 '18 at 12:23
avan1235avan1235
3578
edited Dec 23 '18 at 12:30
Blue
49.3k870157
asked Dec 23 '18 at 12:23
avan1235avan1235
3578
edited Dec 23 '18 at 12:30
Blue
49.3k870157
edited Dec 23 '18 at 12:30
Blue
49.3k870157
edited Dec 23 '18 at 12:30
Blue
49.3k870157
49.3k870157
asked Dec 23 '18 at 12:23
avan1235avan1235
3578
asked Dec 23 '18 at 12:23
avan1235avan1235
3578
asked Dec 23 '18 at 12:23
avan1235avan1235
3578
3578
5
$begingroup$
If I understand correctly, by MERTENS' Theorem, $S$ is equal to the product of the of the initial sums, right? The first is $-ln(2)$ and the second is $frac{3}{2}$ by the geometric series.
$endgroup$
– Viktor Glombik
Dec 23 '18 at 12:38
1
$begingroup$
Yes but how to find it using only the definition of Cauchy product of series?
$endgroup$
– avan1235
Dec 23 '18 at 12:44
$begingroup$
If you were presented with this sum as a $S$ and asked to evaluate it. I would think that decomposing it (via cauchy product) would be the easiest way of coming to a conclusion.
$endgroup$
– Mason
Dec 23 '18 at 16:11
add a comment |
5
$begingroup$
If I understand correctly, by MERTENS' Theorem, $S$ is equal to the product of the of the initial sums, right? The first is $-ln(2)$ and the second is $frac{3}{2}$ by the geometric series.
$endgroup$
– Viktor Glombik
Dec 23 '18 at 12:38
1
$begingroup$
Yes but how to find it using only the definition of Cauchy product of series?
$endgroup$
– avan1235
Dec 23 '18 at 12:44
$begingroup$
If you were presented with this sum as a $S$ and asked to evaluate it. I would think that decomposing it (via cauchy product) would be the easiest way of coming to a conclusion.
$endgroup$
– Mason
Dec 23 '18 at 16:11
5
5
$begingroup$
If I understand correctly, by MERTENS' Theorem, $S$ is equal to the product of the of the initial sums, right? The first is $-ln(2)$ and the second is $frac{3}{2}$ by the geometric series.
$endgroup$
– Viktor Glombik
Dec 23 '18 at 12:38
$begingroup$
If I understand correctly, by MERTENS' Theorem, $S$ is equal to the product of the of the initial sums, right? The first is $-ln(2)$ and the second is $frac{3}{2}$ by the geometric series.
$endgroup$
– Viktor Glombik
Dec 23 '18 at 12:38
1
1
$begingroup$
Yes but how to find it using only the definition of Cauchy product of series?
$endgroup$
– avan1235
Dec 23 '18 at 12:44
$begingroup$
Yes but how to find it using only the definition of Cauchy product of series?
$endgroup$
– avan1235
Dec 23 '18 at 12:44
$begingroup$
If you were presented with this sum as a $S$ and asked to evaluate it. I would think that decomposing it (via cauchy product) would be the easiest way of coming to a conclusion.
$endgroup$
– Mason
Dec 23 '18 at 16:11
$begingroup$
If you were presented with this sum as a $S$ and asked to evaluate it. I would think that decomposing it (via cauchy product) would be the easiest way of coming to a conclusion.
$endgroup$
– Mason
Dec 23 '18 at 16:11
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here is my "lengthy" solution (elaborating @ViktorGlombik comment). Let's start with your sum:
$$S=sum_{n=0}^inftyfrac{1}{3^n}sum_{k=0}^nfrac{1}{3^{-k}}frac{(-1)^{k+1}}{k+1}$$
where
$$sum_{k=0}^nfrac{1}{3^{-k}}frac{(-1)^{k+1}}{k+1}=(-3)^{n+1}Phi(-3,1,n+2)-frac{2}{3}log(2)$$
where $Phi(a,b,c)$ is so-called Lerch transcendent.
Also there is one useful identity: $Phi(-3, 1, n+2) = frac{1}{Gamma(1)}int_0^inftyfrac{e^{-(n+2)t}}{1+3e^{-t}}dt$.
By substitution $u=1+3e^{-t}$ (and $Gamma(1) = 1$), this integral leads to
$$Phi(-3, 1, n+2) = frac{1}{3^{n+2}}int_1^4 frac{(u-1)^{n+1}}{u} du.$$
Plugging it back (observe $uin(1,4)$ for the convergence of the geometric series), we obtain:
$$S=sum_{n=0}^inftyfrac{1}{3^n}left[(-3)^{n+1}frac{1}{3^{n+2}}int_1^4frac{(u-1)^{n+1}}{u}du-frac{2}{3}log(2)right]=$$
$$=sum_{n=0}^inftyfrac{1}{3^{n+1}}int_1^4frac{(1-u)^{n+1}}{u}du-log(2)=$$
$$=int_1^4frac{1}{u}sum_{n=0}^{infty}left(frac{1-u}{3}right)^{n+1}du - log(2)=$$
$$=int_1^4frac{1}{u}frac{1-u}{u+2}du-log(2)=-frac{3}{2}log(2)+log(2)-log(2)=-frac{3}{2}log(2).$$
So yes, indeed the sum converges to $S=-frac{3}{2}log(2)$.
answered Dec 23 '18 at 22:15
pisoirpisoir
643721
$endgroup$
add a comment |
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
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%2f3050303%2fevaluating-the-cauchy-product-of-sum-n-0-infty-frac-1n1n1-and%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is my "lengthy" solution (elaborating @ViktorGlombik comment). Let's start with your sum:
$$S=sum_{n=0}^inftyfrac{1}{3^n}sum_{k=0}^nfrac{1}{3^{-k}}frac{(-1)^{k+1}}{k+1}$$
where
$$sum_{k=0}^nfrac{1}{3^{-k}}frac{(-1)^{k+1}}{k+1}=(-3)^{n+1}Phi(-3,1,n+2)-frac{2}{3}log(2)$$
where $Phi(a,b,c)$ is so-called Lerch transcendent.
Also there is one useful identity: $Phi(-3, 1, n+2) = frac{1}{Gamma(1)}int_0^inftyfrac{e^{-(n+2)t}}{1+3e^{-t}}dt$.
By substitution $u=1+3e^{-t}$ (and $Gamma(1) = 1$), this integral leads to
$$Phi(-3, 1, n+2) = frac{1}{3^{n+2}}int_1^4 frac{(u-1)^{n+1}}{u} du.$$
Plugging it back (observe $uin(1,4)$ for the convergence of the geometric series), we obtain:
$$S=sum_{n=0}^inftyfrac{1}{3^n}left[(-3)^{n+1}frac{1}{3^{n+2}}int_1^4frac{(u-1)^{n+1}}{u}du-frac{2}{3}log(2)right]=$$
$$=sum_{n=0}^inftyfrac{1}{3^{n+1}}int_1^4frac{(1-u)^{n+1}}{u}du-log(2)=$$
$$=int_1^4frac{1}{u}sum_{n=0}^{infty}left(frac{1-u}{3}right)^{n+1}du - log(2)=$$
$$=int_1^4frac{1}{u}frac{1-u}{u+2}du-log(2)=-frac{3}{2}log(2)+log(2)-log(2)=-frac{3}{2}log(2).$$
So yes, indeed the sum converges to $S=-frac{3}{2}log(2)$.
answered Dec 23 '18 at 22:15
pisoirpisoir
643721
$endgroup$
add a comment |
$begingroup$
Here is my "lengthy" solution (elaborating @ViktorGlombik comment). Let's start with your sum:
$$S=sum_{n=0}^inftyfrac{1}{3^n}sum_{k=0}^nfrac{1}{3^{-k}}frac{(-1)^{k+1}}{k+1}$$
where
$$sum_{k=0}^nfrac{1}{3^{-k}}frac{(-1)^{k+1}}{k+1}=(-3)^{n+1}Phi(-3,1,n+2)-frac{2}{3}log(2)$$
where $Phi(a,b,c)$ is so-called Lerch transcendent.
Also there is one useful identity: $Phi(-3, 1, n+2) = frac{1}{Gamma(1)}int_0^inftyfrac{e^{-(n+2)t}}{1+3e^{-t}}dt$.
By substitution $u=1+3e^{-t}$ (and $Gamma(1) = 1$), this integral leads to
$$Phi(-3, 1, n+2) = frac{1}{3^{n+2}}int_1^4 frac{(u-1)^{n+1}}{u} du.$$
Plugging it back (observe $uin(1,4)$ for the convergence of the geometric series), we obtain:
$$S=sum_{n=0}^inftyfrac{1}{3^n}left[(-3)^{n+1}frac{1}{3^{n+2}}int_1^4frac{(u-1)^{n+1}}{u}du-frac{2}{3}log(2)right]=$$
$$=sum_{n=0}^inftyfrac{1}{3^{n+1}}int_1^4frac{(1-u)^{n+1}}{u}du-log(2)=$$
$$=int_1^4frac{1}{u}sum_{n=0}^{infty}left(frac{1-u}{3}right)^{n+1}du - log(2)=$$
$$=int_1^4frac{1}{u}frac{1-u}{u+2}du-log(2)=-frac{3}{2}log(2)+log(2)-log(2)=-frac{3}{2}log(2).$$
So yes, indeed the sum converges to $S=-frac{3}{2}log(2)$.
answered Dec 23 '18 at 22:15
pisoirpisoir
643721
$endgroup$
add a comment |
$begingroup$
Here is my "lengthy" solution (elaborating @ViktorGlombik comment). Let's start with your sum:
$$S=sum_{n=0}^inftyfrac{1}{3^n}sum_{k=0}^nfrac{1}{3^{-k}}frac{(-1)^{k+1}}{k+1}$$
where
$$sum_{k=0}^nfrac{1}{3^{-k}}frac{(-1)^{k+1}}{k+1}=(-3)^{n+1}Phi(-3,1,n+2)-frac{2}{3}log(2)$$
where $Phi(a,b,c)$ is so-called Lerch transcendent.
Also there is one useful identity: $Phi(-3, 1, n+2) = frac{1}{Gamma(1)}int_0^inftyfrac{e^{-(n+2)t}}{1+3e^{-t}}dt$.
By substitution $u=1+3e^{-t}$ (and $Gamma(1) = 1$), this integral leads to
$$Phi(-3, 1, n+2) = frac{1}{3^{n+2}}int_1^4 frac{(u-1)^{n+1}}{u} du.$$
Plugging it back (observe $uin(1,4)$ for the convergence of the geometric series), we obtain:
$$S=sum_{n=0}^inftyfrac{1}{3^n}left[(-3)^{n+1}frac{1}{3^{n+2}}int_1^4frac{(u-1)^{n+1}}{u}du-frac{2}{3}log(2)right]=$$
$$=sum_{n=0}^inftyfrac{1}{3^{n+1}}int_1^4frac{(1-u)^{n+1}}{u}du-log(2)=$$
$$=int_1^4frac{1}{u}sum_{n=0}^{infty}left(frac{1-u}{3}right)^{n+1}du - log(2)=$$
$$=int_1^4frac{1}{u}frac{1-u}{u+2}du-log(2)=-frac{3}{2}log(2)+log(2)-log(2)=-frac{3}{2}log(2).$$
So yes, indeed the sum converges to $S=-frac{3}{2}log(2)$.
answered Dec 23 '18 at 22:15
pisoirpisoir
643721
$endgroup$
Here is my "lengthy" solution (elaborating @ViktorGlombik comment). Let's start with your sum:
$$S=sum_{n=0}^inftyfrac{1}{3^n}sum_{k=0}^nfrac{1}{3^{-k}}frac{(-1)^{k+1}}{k+1}$$
where
$$sum_{k=0}^nfrac{1}{3^{-k}}frac{(-1)^{k+1}}{k+1}=(-3)^{n+1}Phi(-3,1,n+2)-frac{2}{3}log(2)$$
where $Phi(a,b,c)$ is so-called Lerch transcendent.
Also there is one useful identity: $Phi(-3, 1, n+2) = frac{1}{Gamma(1)}int_0^inftyfrac{e^{-(n+2)t}}{1+3e^{-t}}dt$.
By substitution $u=1+3e^{-t}$ (and $Gamma(1) = 1$), this integral leads to
$$Phi(-3, 1, n+2) = frac{1}{3^{n+2}}int_1^4 frac{(u-1)^{n+1}}{u} du.$$
Plugging it back (observe $uin(1,4)$ for the convergence of the geometric series), we obtain:
$$S=sum_{n=0}^inftyfrac{1}{3^n}left[(-3)^{n+1}frac{1}{3^{n+2}}int_1^4frac{(u-1)^{n+1}}{u}du-frac{2}{3}log(2)right]=$$
$$=sum_{n=0}^inftyfrac{1}{3^{n+1}}int_1^4frac{(1-u)^{n+1}}{u}du-log(2)=$$
$$=int_1^4frac{1}{u}sum_{n=0}^{infty}left(frac{1-u}{3}right)^{n+1}du - log(2)=$$
$$=int_1^4frac{1}{u}frac{1-u}{u+2}du-log(2)=-frac{3}{2}log(2)+log(2)-log(2)=-frac{3}{2}log(2).$$
So yes, indeed the sum converges to $S=-frac{3}{2}log(2)$.
answered Dec 23 '18 at 22:15
pisoirpisoir
643721
edited Dec 25 '18 at 9:18
answered Dec 23 '18 at 22:15
pisoirpisoir
643721
answered Dec 23 '18 at 22:15
pisoirpisoir
643721
answered Dec 23 '18 at 22:15
pisoirpisoir
643721
643721
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%2f3050303%2fevaluating-the-cauchy-product-of-sum-n-0-infty-frac-1n1n1-and%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
5
$begingroup$
If I understand correctly, by MERTENS' Theorem, $S$ is equal to the product of the of the initial sums, right? The first is $-ln(2)$ and the second is $frac{3}{2}$ by the geometric series.
$endgroup$
– Viktor Glombik
Dec 23 '18 at 12:38
1
$begingroup$
Yes but how to find it using only the definition of Cauchy product of series?
$endgroup$
– avan1235
Dec 23 '18 at 12:44
$begingroup$
If you were presented with this sum as a $S$ and asked to evaluate it. I would think that decomposing it (via cauchy product) would be the easiest way of coming to a conclusion.
$endgroup$
– Mason
Dec 23 '18 at 16:11