Why is it so hard to find a closed form for $sum_{n=1}^infty frac{1}{n^3}$?
$begingroup$
The series $sum_{n=1}^infty frac{1}{n^p}$ converges for $p>1$; I have known this result since I took calculus in my freshman year. It is also known that $$sum_{n=1}^infty frac{1}{n^2}=frac{pi^2}{6} text{ and } sum_{n=1}^infty frac{1}{n^4}=frac{pi^4}{90}$$
I learned a few years later--while taking a history and philosophy of mathematics course of all places--that the precise value for which the series converges for $p=3$ is still unknown. Doing some brief investigation--using Wolfram|Alpha and Wikipedia--it appears that the result is defined in terms of the Riemann zeta function, i.e. $$sum_{n=1}^infty frac{1}{n^3}=zeta(3)approx 1.2020569...$$
I have been told that a closed form has not been found. My question is why the mechanisms we have developed that enable us to find a closed form in the previous two cases fail in the third case? What it is that makes the $p=3$ case much more difficult? Is there even a closed form for the $p=3$ case, and for that matter, for all odd-numbered cases? I have attempted to ask one of my math professors this question, but I did not quite understand the explanation at the time.
For added context, I am currently doing an undergraduate degree in statistics. As such, I have only taken two semesters of real analysis, and one semester of complex analysis. I also feel like this question might have been have been asked on this site before; if so, please point me in the right direction.
calculus sequences-and-series complex-analysis
$endgroup$
|
show 4 more comments
$begingroup$
The series $sum_{n=1}^infty frac{1}{n^p}$ converges for $p>1$; I have known this result since I took calculus in my freshman year. It is also known that $$sum_{n=1}^infty frac{1}{n^2}=frac{pi^2}{6} text{ and } sum_{n=1}^infty frac{1}{n^4}=frac{pi^4}{90}$$
I learned a few years later--while taking a history and philosophy of mathematics course of all places--that the precise value for which the series converges for $p=3$ is still unknown. Doing some brief investigation--using Wolfram|Alpha and Wikipedia--it appears that the result is defined in terms of the Riemann zeta function, i.e. $$sum_{n=1}^infty frac{1}{n^3}=zeta(3)approx 1.2020569...$$
I have been told that a closed form has not been found. My question is why the mechanisms we have developed that enable us to find a closed form in the previous two cases fail in the third case? What it is that makes the $p=3$ case much more difficult? Is there even a closed form for the $p=3$ case, and for that matter, for all odd-numbered cases? I have attempted to ask one of my math professors this question, but I did not quite understand the explanation at the time.
For added context, I am currently doing an undergraduate degree in statistics. As such, I have only taken two semesters of real analysis, and one semester of complex analysis. I also feel like this question might have been have been asked on this site before; if so, please point me in the right direction.
calculus sequences-and-series complex-analysis
$endgroup$
$begingroup$
Search "Apery" on this site. Your exact question was asked by someone. I will point to it if I find it.
$endgroup$
– Srini
Dec 30 '16 at 3:52
17
$begingroup$
The question is somewhat backwards: what is surprising is that one can find closed forms for the other series! There is no reason at all for a series, no matter how nice the general term, has a sum with a nice closed form.
$endgroup$
– Mariano Suárez-Álvarez
Dec 30 '16 at 3:52
$begingroup$
That one can solve the even cases can be traced back to the fact that when you plug $pi$ into the Taylor expansion of $sin(x)$ you should get 0. There is nothing similar for the odd cases.
$endgroup$
– Fabian
Dec 30 '16 at 4:00
$begingroup$
This thread may help you.
$endgroup$
– Rohan
Dec 30 '16 at 4:11
2
$begingroup$
See this similar question we can compute a closed-form of $zeta(2n)$ because it is the derivative of $cos(z)/sin(z)$ at $z=1/2$, but $zeta(2n+1)$ is the derivative of $Gamma'(z)/Gamma(z)$ at $z=1/2$ which is much more complicated
$endgroup$
– reuns
Dec 30 '16 at 10:29
|
show 4 more comments
$begingroup$
The series $sum_{n=1}^infty frac{1}{n^p}$ converges for $p>1$; I have known this result since I took calculus in my freshman year. It is also known that $$sum_{n=1}^infty frac{1}{n^2}=frac{pi^2}{6} text{ and } sum_{n=1}^infty frac{1}{n^4}=frac{pi^4}{90}$$
I learned a few years later--while taking a history and philosophy of mathematics course of all places--that the precise value for which the series converges for $p=3$ is still unknown. Doing some brief investigation--using Wolfram|Alpha and Wikipedia--it appears that the result is defined in terms of the Riemann zeta function, i.e. $$sum_{n=1}^infty frac{1}{n^3}=zeta(3)approx 1.2020569...$$
I have been told that a closed form has not been found. My question is why the mechanisms we have developed that enable us to find a closed form in the previous two cases fail in the third case? What it is that makes the $p=3$ case much more difficult? Is there even a closed form for the $p=3$ case, and for that matter, for all odd-numbered cases? I have attempted to ask one of my math professors this question, but I did not quite understand the explanation at the time.
For added context, I am currently doing an undergraduate degree in statistics. As such, I have only taken two semesters of real analysis, and one semester of complex analysis. I also feel like this question might have been have been asked on this site before; if so, please point me in the right direction.
calculus sequences-and-series complex-analysis
$endgroup$
The series $sum_{n=1}^infty frac{1}{n^p}$ converges for $p>1$; I have known this result since I took calculus in my freshman year. It is also known that $$sum_{n=1}^infty frac{1}{n^2}=frac{pi^2}{6} text{ and } sum_{n=1}^infty frac{1}{n^4}=frac{pi^4}{90}$$
I learned a few years later--while taking a history and philosophy of mathematics course of all places--that the precise value for which the series converges for $p=3$ is still unknown. Doing some brief investigation--using Wolfram|Alpha and Wikipedia--it appears that the result is defined in terms of the Riemann zeta function, i.e. $$sum_{n=1}^infty frac{1}{n^3}=zeta(3)approx 1.2020569...$$
I have been told that a closed form has not been found. My question is why the mechanisms we have developed that enable us to find a closed form in the previous two cases fail in the third case? What it is that makes the $p=3$ case much more difficult? Is there even a closed form for the $p=3$ case, and for that matter, for all odd-numbered cases? I have attempted to ask one of my math professors this question, but I did not quite understand the explanation at the time.
For added context, I am currently doing an undergraduate degree in statistics. As such, I have only taken two semesters of real analysis, and one semester of complex analysis. I also feel like this question might have been have been asked on this site before; if so, please point me in the right direction.
calculus sequences-and-series complex-analysis
calculus sequences-and-series complex-analysis
asked Dec 30 '16 at 3:45
Douglas FirDouglas Fir
1,224918
1,224918
$begingroup$
Search "Apery" on this site. Your exact question was asked by someone. I will point to it if I find it.
$endgroup$
– Srini
Dec 30 '16 at 3:52
17
$begingroup$
The question is somewhat backwards: what is surprising is that one can find closed forms for the other series! There is no reason at all for a series, no matter how nice the general term, has a sum with a nice closed form.
$endgroup$
– Mariano Suárez-Álvarez
Dec 30 '16 at 3:52
$begingroup$
That one can solve the even cases can be traced back to the fact that when you plug $pi$ into the Taylor expansion of $sin(x)$ you should get 0. There is nothing similar for the odd cases.
$endgroup$
– Fabian
Dec 30 '16 at 4:00
$begingroup$
This thread may help you.
$endgroup$
– Rohan
Dec 30 '16 at 4:11
2
$begingroup$
See this similar question we can compute a closed-form of $zeta(2n)$ because it is the derivative of $cos(z)/sin(z)$ at $z=1/2$, but $zeta(2n+1)$ is the derivative of $Gamma'(z)/Gamma(z)$ at $z=1/2$ which is much more complicated
$endgroup$
– reuns
Dec 30 '16 at 10:29
|
show 4 more comments
$begingroup$
Search "Apery" on this site. Your exact question was asked by someone. I will point to it if I find it.
$endgroup$
– Srini
Dec 30 '16 at 3:52
17
$begingroup$
The question is somewhat backwards: what is surprising is that one can find closed forms for the other series! There is no reason at all for a series, no matter how nice the general term, has a sum with a nice closed form.
$endgroup$
– Mariano Suárez-Álvarez
Dec 30 '16 at 3:52
$begingroup$
That one can solve the even cases can be traced back to the fact that when you plug $pi$ into the Taylor expansion of $sin(x)$ you should get 0. There is nothing similar for the odd cases.
$endgroup$
– Fabian
Dec 30 '16 at 4:00
$begingroup$
This thread may help you.
$endgroup$
– Rohan
Dec 30 '16 at 4:11
2
$begingroup$
See this similar question we can compute a closed-form of $zeta(2n)$ because it is the derivative of $cos(z)/sin(z)$ at $z=1/2$, but $zeta(2n+1)$ is the derivative of $Gamma'(z)/Gamma(z)$ at $z=1/2$ which is much more complicated
$endgroup$
– reuns
Dec 30 '16 at 10:29
$begingroup$
Search "Apery" on this site. Your exact question was asked by someone. I will point to it if I find it.
$endgroup$
– Srini
Dec 30 '16 at 3:52
$begingroup$
Search "Apery" on this site. Your exact question was asked by someone. I will point to it if I find it.
$endgroup$
– Srini
Dec 30 '16 at 3:52
17
17
$begingroup$
The question is somewhat backwards: what is surprising is that one can find closed forms for the other series! There is no reason at all for a series, no matter how nice the general term, has a sum with a nice closed form.
$endgroup$
– Mariano Suárez-Álvarez
Dec 30 '16 at 3:52
$begingroup$
The question is somewhat backwards: what is surprising is that one can find closed forms for the other series! There is no reason at all for a series, no matter how nice the general term, has a sum with a nice closed form.
$endgroup$
– Mariano Suárez-Álvarez
Dec 30 '16 at 3:52
$begingroup$
That one can solve the even cases can be traced back to the fact that when you plug $pi$ into the Taylor expansion of $sin(x)$ you should get 0. There is nothing similar for the odd cases.
$endgroup$
– Fabian
Dec 30 '16 at 4:00
$begingroup$
That one can solve the even cases can be traced back to the fact that when you plug $pi$ into the Taylor expansion of $sin(x)$ you should get 0. There is nothing similar for the odd cases.
$endgroup$
– Fabian
Dec 30 '16 at 4:00
$begingroup$
This thread may help you.
$endgroup$
– Rohan
Dec 30 '16 at 4:11
$begingroup$
This thread may help you.
$endgroup$
– Rohan
Dec 30 '16 at 4:11
2
2
$begingroup$
See this similar question we can compute a closed-form of $zeta(2n)$ because it is the derivative of $cos(z)/sin(z)$ at $z=1/2$, but $zeta(2n+1)$ is the derivative of $Gamma'(z)/Gamma(z)$ at $z=1/2$ which is much more complicated
$endgroup$
– reuns
Dec 30 '16 at 10:29
$begingroup$
See this similar question we can compute a closed-form of $zeta(2n)$ because it is the derivative of $cos(z)/sin(z)$ at $z=1/2$, but $zeta(2n+1)$ is the derivative of $Gamma'(z)/Gamma(z)$ at $z=1/2$ which is much more complicated
$endgroup$
– reuns
Dec 30 '16 at 10:29
|
show 4 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Roger Apéry proved in 1978 that $zeta(3)=displaystyle sum_{n=1}^{infty} n^{-3}$ is irrational, however the irrationality of $displaystylefrac{zeta(3)}{pi^3}$ is still an open problem. No closed form for its value is currently known.
There are however some formulas expressing $zeta(3)$ (and other odd zeta values) in terms of powers of $pi$, but these are not closed forms. The most well known ones are due to Plouffe and Borwein & Bradley:
$$
begin{aligned}
zeta(3)&=frac{7pi^3}{180}-2sum_{n=1}^infty frac{1}{n^3(e^{2pi n}-1)},\
sum_{n=1}^infty frac{1}{n^3,binom {2n}n} &= -frac{4}{3},zeta(3)+frac{pisqrt{3}}{2cdot 3^2},left(zeta(2, tfrac{1}{3})-zeta(2,tfrac{2}{3}) right).
end{aligned}
$$
Moreover, in this Math.SE post we have:
$$
frac{3}{2},zeta(3) = frac{pi^3}{24}sqrt{2}-2sum_{k=1}^infty frac{1}{k^3(e^{pi ksqrt{2}}-1)}-sum_{k=1}^inftyfrac{1}{k^3(e^{2pi ksqrt{2}}-1)}.
$$
You can also check out this paper by Vepstas, which provides a nice generalization to some of these identities.
$endgroup$
add a comment |
$begingroup$
It can be shown that $frac{sin pi x}{pi x}=prod_{kge 1}(1-frac{x^2}{k^2})$. The left-hand side's Maclaurin series can be used to write $pi^{2n}$ in terms of $zeta(2k),,1le kle n$. There is no similar theoretical treatment of $zeta(m)$ for odd $mge 3$.
$endgroup$
$begingroup$
An interesting attempt to extend Euler's method to $zeta(3)$ can be found here: math.stackexchange.com/questions/1459709/…
$endgroup$
– Klangen
Dec 13 '18 at 9:48
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
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active
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votes
$begingroup$
Roger Apéry proved in 1978 that $zeta(3)=displaystyle sum_{n=1}^{infty} n^{-3}$ is irrational, however the irrationality of $displaystylefrac{zeta(3)}{pi^3}$ is still an open problem. No closed form for its value is currently known.
There are however some formulas expressing $zeta(3)$ (and other odd zeta values) in terms of powers of $pi$, but these are not closed forms. The most well known ones are due to Plouffe and Borwein & Bradley:
$$
begin{aligned}
zeta(3)&=frac{7pi^3}{180}-2sum_{n=1}^infty frac{1}{n^3(e^{2pi n}-1)},\
sum_{n=1}^infty frac{1}{n^3,binom {2n}n} &= -frac{4}{3},zeta(3)+frac{pisqrt{3}}{2cdot 3^2},left(zeta(2, tfrac{1}{3})-zeta(2,tfrac{2}{3}) right).
end{aligned}
$$
Moreover, in this Math.SE post we have:
$$
frac{3}{2},zeta(3) = frac{pi^3}{24}sqrt{2}-2sum_{k=1}^infty frac{1}{k^3(e^{pi ksqrt{2}}-1)}-sum_{k=1}^inftyfrac{1}{k^3(e^{2pi ksqrt{2}}-1)}.
$$
You can also check out this paper by Vepstas, which provides a nice generalization to some of these identities.
$endgroup$
add a comment |
$begingroup$
Roger Apéry proved in 1978 that $zeta(3)=displaystyle sum_{n=1}^{infty} n^{-3}$ is irrational, however the irrationality of $displaystylefrac{zeta(3)}{pi^3}$ is still an open problem. No closed form for its value is currently known.
There are however some formulas expressing $zeta(3)$ (and other odd zeta values) in terms of powers of $pi$, but these are not closed forms. The most well known ones are due to Plouffe and Borwein & Bradley:
$$
begin{aligned}
zeta(3)&=frac{7pi^3}{180}-2sum_{n=1}^infty frac{1}{n^3(e^{2pi n}-1)},\
sum_{n=1}^infty frac{1}{n^3,binom {2n}n} &= -frac{4}{3},zeta(3)+frac{pisqrt{3}}{2cdot 3^2},left(zeta(2, tfrac{1}{3})-zeta(2,tfrac{2}{3}) right).
end{aligned}
$$
Moreover, in this Math.SE post we have:
$$
frac{3}{2},zeta(3) = frac{pi^3}{24}sqrt{2}-2sum_{k=1}^infty frac{1}{k^3(e^{pi ksqrt{2}}-1)}-sum_{k=1}^inftyfrac{1}{k^3(e^{2pi ksqrt{2}}-1)}.
$$
You can also check out this paper by Vepstas, which provides a nice generalization to some of these identities.
$endgroup$
add a comment |
$begingroup$
Roger Apéry proved in 1978 that $zeta(3)=displaystyle sum_{n=1}^{infty} n^{-3}$ is irrational, however the irrationality of $displaystylefrac{zeta(3)}{pi^3}$ is still an open problem. No closed form for its value is currently known.
There are however some formulas expressing $zeta(3)$ (and other odd zeta values) in terms of powers of $pi$, but these are not closed forms. The most well known ones are due to Plouffe and Borwein & Bradley:
$$
begin{aligned}
zeta(3)&=frac{7pi^3}{180}-2sum_{n=1}^infty frac{1}{n^3(e^{2pi n}-1)},\
sum_{n=1}^infty frac{1}{n^3,binom {2n}n} &= -frac{4}{3},zeta(3)+frac{pisqrt{3}}{2cdot 3^2},left(zeta(2, tfrac{1}{3})-zeta(2,tfrac{2}{3}) right).
end{aligned}
$$
Moreover, in this Math.SE post we have:
$$
frac{3}{2},zeta(3) = frac{pi^3}{24}sqrt{2}-2sum_{k=1}^infty frac{1}{k^3(e^{pi ksqrt{2}}-1)}-sum_{k=1}^inftyfrac{1}{k^3(e^{2pi ksqrt{2}}-1)}.
$$
You can also check out this paper by Vepstas, which provides a nice generalization to some of these identities.
$endgroup$
Roger Apéry proved in 1978 that $zeta(3)=displaystyle sum_{n=1}^{infty} n^{-3}$ is irrational, however the irrationality of $displaystylefrac{zeta(3)}{pi^3}$ is still an open problem. No closed form for its value is currently known.
There are however some formulas expressing $zeta(3)$ (and other odd zeta values) in terms of powers of $pi$, but these are not closed forms. The most well known ones are due to Plouffe and Borwein & Bradley:
$$
begin{aligned}
zeta(3)&=frac{7pi^3}{180}-2sum_{n=1}^infty frac{1}{n^3(e^{2pi n}-1)},\
sum_{n=1}^infty frac{1}{n^3,binom {2n}n} &= -frac{4}{3},zeta(3)+frac{pisqrt{3}}{2cdot 3^2},left(zeta(2, tfrac{1}{3})-zeta(2,tfrac{2}{3}) right).
end{aligned}
$$
Moreover, in this Math.SE post we have:
$$
frac{3}{2},zeta(3) = frac{pi^3}{24}sqrt{2}-2sum_{k=1}^infty frac{1}{k^3(e^{pi ksqrt{2}}-1)}-sum_{k=1}^inftyfrac{1}{k^3(e^{2pi ksqrt{2}}-1)}.
$$
You can also check out this paper by Vepstas, which provides a nice generalization to some of these identities.
edited Dec 13 '18 at 9:08
answered Dec 13 '18 at 8:58
KlangenKlangen
1,75811334
1,75811334
add a comment |
add a comment |
$begingroup$
It can be shown that $frac{sin pi x}{pi x}=prod_{kge 1}(1-frac{x^2}{k^2})$. The left-hand side's Maclaurin series can be used to write $pi^{2n}$ in terms of $zeta(2k),,1le kle n$. There is no similar theoretical treatment of $zeta(m)$ for odd $mge 3$.
$endgroup$
$begingroup$
An interesting attempt to extend Euler's method to $zeta(3)$ can be found here: math.stackexchange.com/questions/1459709/…
$endgroup$
– Klangen
Dec 13 '18 at 9:48
add a comment |
$begingroup$
It can be shown that $frac{sin pi x}{pi x}=prod_{kge 1}(1-frac{x^2}{k^2})$. The left-hand side's Maclaurin series can be used to write $pi^{2n}$ in terms of $zeta(2k),,1le kle n$. There is no similar theoretical treatment of $zeta(m)$ for odd $mge 3$.
$endgroup$
$begingroup$
An interesting attempt to extend Euler's method to $zeta(3)$ can be found here: math.stackexchange.com/questions/1459709/…
$endgroup$
– Klangen
Dec 13 '18 at 9:48
add a comment |
$begingroup$
It can be shown that $frac{sin pi x}{pi x}=prod_{kge 1}(1-frac{x^2}{k^2})$. The left-hand side's Maclaurin series can be used to write $pi^{2n}$ in terms of $zeta(2k),,1le kle n$. There is no similar theoretical treatment of $zeta(m)$ for odd $mge 3$.
$endgroup$
It can be shown that $frac{sin pi x}{pi x}=prod_{kge 1}(1-frac{x^2}{k^2})$. The left-hand side's Maclaurin series can be used to write $pi^{2n}$ in terms of $zeta(2k),,1le kle n$. There is no similar theoretical treatment of $zeta(m)$ for odd $mge 3$.
answered Dec 13 '18 at 9:18
J.G.J.G.
28k22844
28k22844
$begingroup$
An interesting attempt to extend Euler's method to $zeta(3)$ can be found here: math.stackexchange.com/questions/1459709/…
$endgroup$
– Klangen
Dec 13 '18 at 9:48
add a comment |
$begingroup$
An interesting attempt to extend Euler's method to $zeta(3)$ can be found here: math.stackexchange.com/questions/1459709/…
$endgroup$
– Klangen
Dec 13 '18 at 9:48
$begingroup$
An interesting attempt to extend Euler's method to $zeta(3)$ can be found here: math.stackexchange.com/questions/1459709/…
$endgroup$
– Klangen
Dec 13 '18 at 9:48
$begingroup$
An interesting attempt to extend Euler's method to $zeta(3)$ can be found here: math.stackexchange.com/questions/1459709/…
$endgroup$
– Klangen
Dec 13 '18 at 9:48
add a comment |
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$begingroup$
Search "Apery" on this site. Your exact question was asked by someone. I will point to it if I find it.
$endgroup$
– Srini
Dec 30 '16 at 3:52
17
$begingroup$
The question is somewhat backwards: what is surprising is that one can find closed forms for the other series! There is no reason at all for a series, no matter how nice the general term, has a sum with a nice closed form.
$endgroup$
– Mariano Suárez-Álvarez
Dec 30 '16 at 3:52
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That one can solve the even cases can be traced back to the fact that when you plug $pi$ into the Taylor expansion of $sin(x)$ you should get 0. There is nothing similar for the odd cases.
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– Fabian
Dec 30 '16 at 4:00
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This thread may help you.
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– Rohan
Dec 30 '16 at 4:11
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See this similar question we can compute a closed-form of $zeta(2n)$ because it is the derivative of $cos(z)/sin(z)$ at $z=1/2$, but $zeta(2n+1)$ is the derivative of $Gamma'(z)/Gamma(z)$ at $z=1/2$ which is much more complicated
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– reuns
Dec 30 '16 at 10:29