Nice proofs of $\zeta(4) = \frac{\pi^4}{90}$?

Question

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 $\zeta(2) = \pi^2/6$.

Answer 1

In the same spirit of the 1st proof of this answer. If we substitute $\pi $ for $ x $ 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

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

Hence

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

Answer 2

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 $\zeta(4)={\pi^4\over 90}$.

Answer 3

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 $\zeta(6)$.

Answer 4

I would like to present you a method based on Daners' derivation of $\zeta(2)$ (which I summarised on MathStackexchange in here couple years ago).

We start defining (for all $n \in \mathbb{N}_0$)

\begin{align} A_n & =\int_0^{\pi/2}\cos^{2n}x\;\mathrm{d}x,\\ B_n& =\int_0^{\pi/2}x^2\cos^{2n}x\;\mathrm{d}x,\\ C_n& =\int_0^{\pi/2}x^4\cos^{2n}x\;\mathrm{d}x \end{align}

and let $\beta_n = B_n/A_n$ and $\gamma_n = C_n/A_n$.

The first integral for $A_n$ is well known and follows a reccurence relation

$$A_{n}=\frac{2n-1}{2n}A_{n-1}\tag{1}.$$

By applying per partes on the $A_n$ integral twice, we get:

$$A_n=\int_0^{\pi/2}\cos^{2n}x\;\mathrm{d}x=x\cos^{2n}x\bigg{|}_0^{\pi/2}-\frac{x^2}{2}(\cos^{2n}x)'\bigg{|}_0^{\pi/2}+\frac{1}{2}\int_0^{\pi/2}x^2(\cos^{2n}x)''\;\mathrm{d}x$$

The first two terms vanish, so only the integral remains and since $(\cos^{2n}x)''=2n(2n-1)\cos^{2n-2}x-4n^2\cos^{2n}x$, we get for $n\geq 1$:

$$A_n=(2n-1)nB_{n-1}-2n^2B_{n}\tag{2}$$

Similarly, applying per partes twice on $B_n$ instead, we get

$$B_n=\frac16 (2n-1)nC_{n-1}-\frac13 n^2C_{n}\tag{3}$$

Inserting $(1)$ into $(2)$ and $(3)$ and rearranging, we get a following recurence reltions

$$\frac{1\cdot 2}{4}\frac{1}{n^2}=\beta_{n-1}-\beta_n\tag{4}$$ $$\frac{3\cdot 4}{4}\frac{\beta_n}{n^2}=\gamma_{n-1}-\gamma_n\tag{5}$$

Summing these up, we get, by the telescoping property

$$\sum_{l=1}^k\frac{1}{2l^2}=\beta_0-\beta_k \qquad \text{and} \qquad \sum_{k=1}^n\frac{3\beta_k}{k^2}=\gamma_0-\gamma_n. \tag{6}$$

Inserting one sum into another (expressing $\beta_k$), we get $$3\beta_0\sum_{k=1}^n\frac{1}{k^2} - \frac32\sum_{k=1}^n\frac{1}{k^2}\sum_{l=1}^k\frac{1}{l^2}=\gamma_0 - \gamma_n. \tag{7}$$

Note that in general $$2\sum_{k=1}^n a_k\sum_{l=1}^k a_l = 2\sum_{1\leq l \leq k \leq n}a_k a_l = \left(\sum_{k=1}^n a_k\right)^2+\sum_{k=1}^n a_k^2,$$

substituing for $a_k = 1/k^2$ into $(7)$ and rewriting $\sum_{k=1}^n 1/k^2$ as $2\beta_0 - 2\beta_n$ everywhere, we get

$$\bbox[10px,#ffd]{\sum_{k=1}^n \frac{1}{k^4} = 4\beta_0^2 - 4\beta_n^2 - \frac{4}{3}\gamma_0 + \frac{4}{3}\gamma_n.}\tag{8}$$

However, using the inequalities $x^4<(\pi/2)^2 x^2$ and $\frac{2x}{\pi}<\sin x$ valid on $(0,\frac{\pi}{2})$ and by $(1)$, we get

$$0<\gamma_{n-1} <\left(\frac{\pi}{2}\right)^2\beta_{n-1} < \frac{\int_0^{\pi/2}\sin^2x\cos^{2n-2}x}{\int_0^{\pi/2}\cos^{2n-2}x}=\frac{A_{n-1} - A_n}{A_n} = \frac{1}{2n}.$$

Applying the squeeze theorem, we get $$\lim_{n\rightarrow \infty} \beta_n = \lim_{n\rightarrow \infty} \gamma_n = 0$$

and hence, taking the limit of $(8)$,

$$\zeta_4 = \lim_{n\rightarrow \infty} \sum_{k=1}^n\frac{1}{k^4} = 4\beta_0^2-\frac{4}{3}\gamma_0 = 4\left(\frac13 \frac{\pi^2}{2^2}\right)^2-\frac{4}{3} \frac{1}{5}\frac{\pi^4}{2^4} = \frac{\pi^4}{90}$$

This finishes the proof.