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I know some nice ways to prove that $\zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \pi^2/6$. For example, see Robin Chapman's list or the answers to the question "Different methods to compute $\sum_{n=1}^{\infty} \frac{1}{n^2}$?"

Are there any nice ways to prove that $$\zeta(4) = \sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}?$$

I already know some proofs that give all values of $\zeta(n)$ for positive even integers $n$ (like #7 on Robin Chapman's list or Qiaochu Yuan's answer in the linked question). I'm not so much interested in those kinds of proofs as I am those that are specifically for $\zeta(4)$.

I would be particularly interested in a proof that isn't an adaption of one that $\zeta(2) = \pi^2/6$.

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I would say that most of the standard proofs for computing $\zeta(2)$ applies to $\zeta(2n)$ too (here $n>0$ is an integer). –  AD. Mar 21 '11 at 20:07
    
@AD: Yeah, that issue is part of what I was trying to get at with my question: Are there any proofs that $\zeta(4) = \pi^4/90$ that aren't just adaptions of a proof that $\zeta(2) = \pi^2/6$? I probably should have made that more explicit. –  Mike Spivey Mar 21 '11 at 22:58
    
I wondering what does the $\zeta$ represent? Is that of any significance or just a variable? –  night owl Feb 27 '12 at 2:40
1  
@nightowl: It refers to the Riemann zeta function. –  Mike Spivey Feb 27 '12 at 3:22

8 Answers 8

up vote 33 down vote accepted

In the same spirit of the 1st proof of this answer. If we substitute $x$ for $\pi $ in the Fourier trigonometric series expansion of $% f(x)=x^{4}$, with $-\pi \leq x\leq \pi $,

$$x^{4}=\frac{1}{5}\pi ^{4}+\sum_{n=1}^{\infty }\frac{8n^{2}\pi ^{2}-48}{n^{4}}\cos n\pi \cdot \cos nx,$$

we obtain

$$\begin{eqnarray*} \pi ^{4} &=&\frac{1}{5}\pi ^{4}+\sum_{n=1}^{\infty }\frac{8n^{2}\pi ^{2}-48}{% n^{4}}\cos ^{2}n\pi \\ &=&\frac{1}{5}\pi ^{4}+8\pi ^{2}\sum_{n=1}^{\infty }\frac{1}{n^{2}}% -48\sum_{n=1}^{\infty }\frac{1}{n^{4}}\text{.} \end{eqnarray*}$$

Hence

$$\sum_{n=1}^{\infty }\frac{1}{n^{4}}=\frac{\pi ^{4}}{48}\left( -1+\frac{1}{5}+ \frac{8}{6}\right) =\frac{\pi ^{4}}{48}\cdot \frac{8}{15}=\frac{1}{90}\pi ^{4}.$$

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Américo Tavares, can this proof be extended to calculate, recursively, values of $\zeta(2n)$ for larger values of $n$? –  Mike Spivey Mar 21 '11 at 18:35
    
Mike Spivey, I think so, but I am not quite sure. One that can be extended for sure is the second proof in my answer math.stackexchange.com/questions/8337/…. –  Américo Tavares Mar 21 '11 at 18:41
    
Thanks! (extra characters) –  Mike Spivey Mar 21 '11 at 18:49
4  
@Mike Spivey, It can. I posted the computation in this Portuguese post problemasteoremas.wordpress.com/2011/05/25/… . I got $$x^{2p}=\frac{\pi ^{2p}}{2p+1}+\frac{2}{\pi }\sum_{n=1}^{\infty }\left( \cos nx\cdot I_{2p}\right) ,$$ where $$I_{2p}=\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x$$ satisfies $$I_{2p}=\frac{2p}{n^{2}}\pi ^{2p-1}\cos n\pi -\frac{2p\left( 2p-1\right) }{% n^{2}}I_{2\left( p-1\right) }\qquad I_{0}=0$$ –  Américo Tavares May 27 '11 at 10:39
    
"Substitute x for π", not the other way around. –  marshaul Apr 10 at 14:25

Consider the function $f(t):=t^2\ \ (-\pi\leq t\leq \pi)$, extended to all of ${\mathbb R}$ periodically with period $2\pi$. Developping $f$ into a Fourier series we get $$t^2 ={\pi^2\over3}+\sum_{k=1}^\infty {4(-1)^k\over k^2}\cos(kt)\qquad(-\pi\leq t\leq \pi).$$ If we put $t:=\pi$ here we easily find $\zeta(2)={\pi^2\over6}$. For $\zeta(4)$ we use Parseval's formula $$\|f\|^2=\sum_{k=-\infty}^\infty |c_k|^2\ .$$ Here $$\|f\|^2={1\over2\pi}\int_{-\pi}^\pi t^4\>dt={\pi^4\over5}$$ and the $c_k$ are the complex Fourier coefficients of $f$. Therefore $c_0={\pi^2\over3}$ and $|c_{\pm k}|^2={1\over4}a_k^2={4\over k^4}$ $\ (k\geq1)$. Putting it all together gives $\zeta(4)={\pi^4\over 90}$.

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Thanks for your answer, Christian. –  Mike Spivey Mar 22 '11 at 17:56
    
I believe this is exercise 14, of Chapter 8 of Rudin's Principles of Mathematical Analysis 3E. (Page 199) –  Eric Naslund Mar 29 '11 at 16:21

If you are specially interested only in $\zeta(4)$, the following proof would work but this is an adaptation Euler's idea. The idea is just to mimic Euler's proof for the Basel problem. Euler looks at the function whose zeros are at $\pm \pi, \pm 2 \pi, \pm 3 \pi, \ldots$

To evaluate $\zeta(4)$, we can mimic Euler's idea and look at roots at $\pm \pi, \pm i \pi,\pm 2 \pi, \pm 2 i \pi,\pm 3 \pi, \pm 3 i \pi$.

Let $$p(z) = \left(1 - \left(\frac{z}{i \pi}\right)^4 \right) \times \left(1 - \left(\frac{z}{2 i \pi}\right)^4 \right) \times \left(1 - \left(\frac{z}{3 i \pi}\right)^4 \right) \times \cdots$$

It is not hard to guess that $p(z)$ is same as $$\frac{i \sin(z) \times \sin \left( \frac{z}{i} \right)}{z^2} = \left(1-\frac{z^2}{3!} + \frac{z^4}{5!} -\cdots \right) \times \left(1+\frac{z^2}{3!} + \frac{z^4}{5!} + \cdots \right)$$

Compare the coefficient of $z^4$ to get $$\zeta(4) = \frac{\pi^4}{90}$$

This proof could be extended for any even number to give that $$\zeta(2n) = (-1)^{n+1} \frac{B_{2n} 2^{2n}}{2(2n)!} \pi^{2n} $$

As expected for odd numbers, this doesn't work. For instance for $3$, if you try to work out by looking at $$p(z) = \left(1 - \left(\frac{z}{\omega \pi}\right)^3 \right) \times \left(1 - \left(\frac{z}{2 \omega \pi}\right)^3 \right) \times \left(1 - \left(\frac{z}{3 \omega \pi}\right)^3 \right) \times \cdots$$ where $\omega^3 = 1$ there is an asymmetry since $$\sin(z) \sin \left( \frac{z}{\omega}\right) \sin \left( \frac{z}{\omega^2}\right)$$ extends on both sides and the non-zero roots are at $$\pm \pi,\pm \omega \pi,\pm \omega^2 \pi,\pm 2 \pi,\pm 2 \omega \pi,\pm 2 \omega^2 \pi,\pm 3 \pi,\pm 3 \omega \pi,\pm 3 \omega^2 \pi,\ldots$$ and hence the $\zeta(3)$ terms nicely hides by canceling out and the resulting expression only gives $\zeta(6)$.

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Thanks, Sivaram. –  Mike Spivey May 27 '11 at 14:02

From How many ways to calculate: $\sum_{n=-\infty}^{+\infty}\frac{1}{(u+n)^2}$ where $u \not \in \Bbb{Z}$, we know that $$ \sum_{n=-\infty}^\infty \frac{1}{(z+n)^2}=\frac{\pi^2}{\sin^2(\pi z)}. $$ Differentiating this twice, we have $$ \sum_{n=-\infty}^\infty \frac{1}{(z+n)^4}=\frac{\pi^4(2+\cos(2\pi x))}{3\sin^4(\pi z)}. $$ So $$ \sum_{n=1}^\infty \left(\frac{1}{(z-n)^4}+\frac{1}{(z+n)^4}\right)=\frac{\pi^4(2+\cos(2\pi x))}{3\sin^4(\pi z)}-\frac{1}{z^4}. $$ Note that the LHS of the above is analytic $z=0$ and hence $$ \sum_{n=1}^\infty\frac{1}{n^4}=\lim_{z\to 0}\frac{1}{2}\left(\frac{\pi^4(2+\cos(2\pi x))}{3\sin^4(\pi z)}-\frac{1}{z^4}\right)=\frac{\pi^4}{90}. $$

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Consider the contour integral $$ \oint_C\frac{\pi\cot\pi z}{z^4}\ dz $$ where $C$ is the counter-clockwise square contour centered at origin with vertices $\left(N+\frac12\right)(\pm1\pm i)$.


Lemma

Suppose that the function $\phi(z)$ is holomorphic at $z=n\in\mathbb{Z}$ with $\phi(n)\neq0$, then $\pi\phi(z)\cot\pi z$ has a simple pole at $n$ with residue $\phi(n)$.

Proof

Note that $\tan\pi z$ have simple zeros at $z=n$, hence $\pi\phi(z)\cot\pi z$ have simple poles there and $$ \text{Res}\left[\pi\phi(z)\cot\pi z\ ;\ n\right]=\text{Res}\left[\frac{\pi\phi(z)}{\tan\pi z}\ ;\ n\right]=\frac{\pi\phi(n)}{\pi\sec^2\pi n}=\phi(n). $$


Thus, by residue theorem for $z\neq0$ we obtain $$ \sum_{n=-N}^N\text{Res}\left[\frac{\pi\cot\pi z}{z^4}\ ;\ z=n\right]=\sum_{n=-N,\ n\neq0}^N\frac1{n^4}=2\sum_{n=1}^N\frac1{n^4}.\tag1 $$ From the Taylor series of $\cot\pi z$ at $z=0$ we obtain \begin{align} \frac{\pi\cot\pi z}{z^4}&=\frac\pi{z^4}\cos\pi z\csc\pi z\\ &=\frac\pi{z^4}\left(1-\frac{(\pi z)^2}{2!}+\frac{(\pi z)^4}{4!}-\frac{(\pi z)^6}{6!}+\cdots\right)\left(\frac1{\pi z}+\frac{\pi z}{6}+\frac{7(\pi z)^3}{360}+\cdots\right)\\ &=\frac1{z^5}\left(1-\frac{(\pi z)^2}{2!}+\frac{(\pi z)^4}{4!}-\frac{(\pi z)^6}{6!}+\cdots\right)\left(1+\frac{(\pi z)^2}{6}+\frac{7(\pi z)^4}{360}+\cdots\right)\\ \end{align} Expanding the series above, we see that $$ \text{Res}\left[\frac{\pi\cot\pi z}{z^4}\ ;\ z=0\right]=-\frac{\pi^4}{2!\cdot6}+\frac{\pi^4}{4!}+\frac{7\pi^4}{360}=-\frac{\pi^4}{45}.\tag2 $$ Observe that at any point on the boundary, we have $$ \left|\frac{\pi\cot\pi z}{z^4}\right|\le\frac{\pi\coth\frac\pi2}{\left(N+\frac12\right)^4}.\tag3 $$


Proof

Putting $z=x+iy$ and using the trigonometric sum formulas and basic identities, we have $$ |\cot\pi z|^2=\left|\frac{\cos\pi z}{\sin\pi z}\right|=\frac{\sinh^2\pi y+\cos^2\pi x}{\cosh^2\pi y-\cos^2\pi x}. $$ On the vertices sides of contour $C$, we have $x=\pm\left(N+\frac12\right)$ giving $\cos\left(N+\frac12\right)\pi=0$, hence $$ |\cot\pi z|=|\tanh\pi y|\le1. $$ On the horizontal sides we have $0\le\cos^2\pi x\le1$, hence $$ |\cot\pi z|^2\le\frac{\sinh^2\pi y+1}{\cosh^2\pi y-1}=\frac{\cosh^2\pi y}{\sinh^2\pi}=\coth^2\pi y. $$ Therefore $$ |\cot\pi z|\le\coth\pi y=\coth\left(N+\frac12\right)\pi\le\coth\frac\pi2 $$ Thus, on the boundary of contour $C$ we have $$ |\cot\pi z|\le\max\left[1,\coth\frac\pi2\right]=\coth\frac\pi2 $$


From $(3)$ and the property $$ \left|\int_C f(z)\ dz\right|\le ML, $$ where $M$ is $\max|f(z)|$ on C and $L$ is the length of $C$, we obtain $$ \oint_C\frac{\pi\cot\pi z}{z^4}\ dz\le\frac{\pi\coth\frac\pi2}{\left(N+\frac12\right)^4}\cdot8\left(N+\frac12\right)=\frac{8\pi\coth\frac\pi2}{\left(N+\frac12\right)^3}\to0 $$ as $N\to\infty$. Thus, using $(1)$ and $(2)$ also using residue theorem we obtain \begin{align} \lim_{N\to\infty}\frac{1}{2\pi i}\oint_C\frac{\pi\cot\pi z}{z^4}\ dz&=0\\ \sum_{n=-\infty}^\infty\text{Res}\left[\frac{\pi\cot\pi z}{z^4}\ ;\ z=n\right]+\text{Res}\left[\frac{\pi\cot\pi z}{z^4}\ ;\ z=0\right]&=0\\ 2\sum_{n=1}^\infty\frac1{n^4}-\frac{\pi^4}{45}&=0\\ \large\color{blue}{\sum_{n=1}^\infty\frac1{n^4}}&\large\color{blue}{=\frac{\pi^4}{90}}. \end{align}

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This is just a sketch of one of a many possible proofs.

Step1. Prove that over the interval $[0,2\pi]$, the function: $$f(x)=\sum_{n=1}^{+\infty}\frac{\cos(nx)}{n^2}$$ is a second degree-polynomial whose graphics passes through the points: $$(0,\pi^2/6),\quad (\pi,-\pi^2/12),\quad (2\pi,\pi^2/6).$$

Step2. Deduce from Lagrange interpolation that: $$ f(x) = \frac{\pi^2}{6}-\frac{x(2\pi-x)}{4}.$$

Step3. Apply Parseval's identity to $f(x)$: $$\int_{0}^{2\pi}f(x)^2\,dx = \pi\sum_{n=1}^{+\infty}\frac{1}{n^4}.$$

Step4. Prove, through the second step, that: $$\int_{0}^{2\pi}f(x)^2\, dx = \frac{\pi^5}{90}.$$

Conclusion:

$$\zeta(4)=\sum_{n=1}^{+\infty}\frac{1}{n^4} = \frac{\pi^4}{90}.$$

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3  
Duplicate of this answer. –  robjohn Aug 7 at 0:26
    
@robjohn: yes, I know, I just copied it from that closed answer to here for visibility, I think it is interesting. –  Jack D'Aurizio Aug 7 at 1:09

By induction we can easily prove that for any nonnegative real numbers $a_k$ $$1-\sum_{k=1}^na_k+\sum_{1\le i<j\le n}a_ia_j-\sum_{1\le i<j<k\le n}a_ia_ja_k\le\prod_{k=1}^n(1-a_k)\le1-\sum_{k=1}^na_k+\sum_{1\le i<j\le n}a_ia_j$$ Taking $a_k=\frac{x^2}{k^2\pi^2}$,we get $$1-\frac{x^2}{\pi^2}\zeta_n(2)+\frac{x^4}{\pi^4}\frac{\zeta_n(2)^2-\zeta_n(4)}2-\frac{x^6}{\pi^6}\frac{\zeta_n(2)^3-\zeta_n(6)}{6}\le\prod_{k=1}^n(1-\frac{x^2}{k^2\pi^2})\le1-\frac{x^2}{\pi^2}\zeta_n(2)+\frac{x^4}{\pi^4}\frac{\zeta_n(2)^2-\zeta_n(4)}2$$ Since $\prod_{k=1}^{\infty}(1-\frac{x^2}{k^2\pi^2})=\frac{\sin(x)}x$(proof here), by taking $n\to\infty$ $$1-\frac{x^2}{\pi^2}\zeta(2)+\frac{x^4}{\pi^4}\frac{\zeta(2)^2-\zeta(4)}2-\frac{x^6}{\pi^6}\frac{\zeta(2)^3-\zeta(6)}{6}\le1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\cdots\le1-\frac{x^2}{\pi^2}\zeta(2)+\frac{x^4}{\pi^4}\frac{\zeta(2)^2-\zeta(4)}2$$ subtraciting $1-\frac{x^2}{\pi^2}\zeta_n(2)$ $$\frac{x^4}{\pi^4}\frac{\zeta(2)^2-\zeta(4)}2-\frac{x^6}{\pi^6}\frac{\zeta(2)^3-\zeta(6)}{6}\le\frac{x^4}{5!}-\frac{x^6}{7!}+\cdots\le\frac{x^4}{\pi^4}\frac{\zeta(2)^2-\zeta(4)}2$$ dividing by $x^4$ and putting $x=0$ we get $$\frac1{\pi^4}\frac{\zeta(2)^2-\zeta(4)}2=\frac1{5!}$$ and this follows that $\zeta(4)=\frac{\pi^4}{90}$.

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Knowing a close form for $\zeta(2)$ there is an algebraic way to get $\zeta(4)$.

Consider $f(m,n):=\dfrac{2}{mn^3}+\dfrac{1}{m^2n^2}+\dfrac{2}{m^3n}$

There is the identity:

$f(m,n)-f(m,n+m)-f(m+n,n)=\dfrac{2}{m^2n^2}$

Therefore:

$\displaystyle \sum_{m,n>0} f(m,n)-\sum_{m,n>0} f(m,n+m)-\sum_{m,n>0} f(m+n,n)=\sum_{m,n>0} \dfrac{2}{m^2n^2}$

The right member is equal to:

$\displaystyle \sum_{m=1}^{+\infty}\Big(\sum_{n=1}^{+\infty}\dfrac{2}{m^2n^2}\Big)=2\sum_{m=1}^{+\infty}\dfrac{1}{n^2}\Big(\sum_{n=1}^{+\infty}\dfrac{1}{m^2}\Big)=2\Big(\sum_{n=1}^{+\infty}\dfrac{1}{n^2}\Big)\Big(\sum_{n=1}^{+\infty}\dfrac{1}{m^2}\Big)=2\zeta(2)^2$

The second sum in the left member is equal to:

$\displaystyle \sum_{n>m>0} f(m,n)$

The third one is equal to:

$\displaystyle \sum_{m>n>0} f(m,n)$

Therefore the left member is equal to:

$\displaystyle \sum_{n=1}^{+\infty}f(n,n)=\sum_{n=1}^{+\infty}\dfrac{5}{n^4}=5\zeta(4)$

(proof found in "Quelques conséquences surprenantes de la cohomologie de $SL_2(\mathbb{Z})$, Don Zagier) http://people.mpim-bonn.mpg.de/zagier/files/tex/ConsequencesCohomologySL/fulltext.pdf

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