# Computing $\sum_{n=1}^{\infty} \frac{1}{n^4}$

I came across this problem in Fourier analysis, but I'm not sure my confusion is with Fourier analysis per se, it might be that I'm lacking some understanding of sums/series.

I am trying to determine the sum $$\sum_{n=1}^{\infty} \frac{1}{n^4}.$$ I know this problem has been posted before, but I want to solve it in a particular way. I already have that $$\sum_{n=0}^{\infty} \frac{1}{(2n+1)^4} = \frac{\pi^4}{96}.$$ I found a suggested solution online that said to solve $x = \frac{\pi^4}{96} + \frac{x}{16}$ for $x$ to find the right answer. (So $x$ is the series I am trying to determine.) But I don't understand why I should divide $x$ by $16$. I understand that I have to add something to $\frac{\pi^4}{96},$ since that is the sum $\sum_{n=1}^{\infty} \frac{1}{n^4}$ over all odd indices, but not over the even ones. And I get why I must add $x$ divided by something, but why $16$?

Hint :$$\sum_{n=0}^{\infty} \frac{1}{(2n+1)^4} +\sum_{n=1}^{\infty} \frac{1}{(2n)^4}=\sum_{n=1}^{\infty} \frac{1}{n^4}$$ and $$\sum_{n=1}^{\infty} \frac{1}{(2n)^4}=\frac1{2^4}\sum_{n=1}^{\infty} \frac{1}{n^4}$$

$$\sum_{n=1}^\infty\frac1{n^4}=\sum_{n=1}^\infty\frac1{(2n)^4}+\sum_{n=0}^\infty\frac1{(2n+1)^4}\implies\frac{15}{16}\sum_{n=1}^\infty\frac1{n^4}=\sum_{n=0}^\infty\frac1{(2n+1)^4}\;\ldots$$

Although the OP is seeking a way forward through Fourier series analysis, I thought it might be instructive and useful for some users to present an approach that relies on contour integration. To that end, we proceed.

We begin by noting that the function $f(z)=\frac{\cot(\pi z)}{z^4}$ has simple poles at $z=n$, $n\ne 0$ and a fifth-order pole at $z=0$. The residue at $z=n$, $n\ne 0$ is given by $\frac1{\pi n^4}$. The residue at $z=0$ is given by $-\frac{\pi^3}{45}$.

Let $C_N$ be the contour $|z|=N+1/2$. Then, we have

\begin{align} \lim_{N\to \infty}\oint_{C_N} \frac{\cot(\pi z)}{z^4}\,dz&=2\pi i \left(2\sum_{n=1}^\infty \frac{1}{\pi n^4} +\text{Res}\left(\frac{\cot(\pi z)}{z^4},z=0\right)\right)\\\\ &=2\pi i \left(\frac{2}{\pi}\sum_{n=1}^\infty \frac{1}{n^4}-\frac{\pi^3}{45}\right)\\\\ &=0 \end{align}

Hence, we see that

$$\bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty \frac{1}{n^4}=\frac{\pi^4}{90}}$$

And we are done!

Sharing ways to do these is always fun, so why not have one more.

We have

$$\sin(z)=z\prod_{n=1}^\infty\left(1-\frac{z^2}{\pi^2n^2}\right)$$

and

$$\sin(z)=\sum_{n=0}^\infty\frac{(-1)^nz^{2n+1}}{(2n+1)!}.$$

Expanding these representations (the sum on the left, the product on the right) gives us

$$z-\frac{z^3}{3!}+\frac{z^5}{5!}-\cdots=z-\frac{z^3}{\pi^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots\right)+\frac{3z^5}{4\pi^4}\left(1+\frac{1}{2^4}+\frac{1}{3^4}+\cdots\right)-\cdots$$

Equating coefficients of like powers gives us

$$\zeta(2)=\frac{\pi^2}{6}\;\text{ and }\;\zeta(4)=\frac{\pi^4}{90}.$$


By 'comparing' two well known $\ds{\cot}$-expansions:

\begin{align} \cot\pars{\pi x} & = {1 \over \pi x} - {1 \over 3}\,\pi x - {1 \over 45}\,\pars{\pi x}^{3} - {2 \over 945}\,\pars{\pi x}^{5} + \,\mrm{O}\pars{x^{7}} \\[5mm] & = \pars{1 \over \pi}{1 \over x} - \pars{\pi \over 3}x - \pars{\color{#f00}{\pi^{3} \over 45}}x^{3} - \pars{{2\pi^{5} \over 945}}x^{5} + \,\mrm{O}\pars{x^{7}} \label{1}\tag{1} \end{align}

\begin{align} \cot\pars{\pi x} & = {1 \over \pi x} + {2 \over \pi}\sum_{n = 1}^{\infty}{x \over x^{2} - n^{2}} = {1 \over \pi x} - {2 \over \pi}\sum_{n = 1}^{\infty}{1 \over n^{2}}{x \over 1 - x^{2}/n^{2}} \\[5mm] & = {1 \over \pi x} - \pars{{2 \over \pi}\,\sum_{n = 1}^{\infty}{1 \over n^{2}}}x - \pars{\color{#f00}{{2 \over \pi}\sum_{n = 1}^{\infty}{1 \over n^{4}}}}x^{3} + \,\mrm{O}\pars{x^{5}} \label{2}\tag{2} \end{align}
With \eqref{1} and \eqref{2}:

$\ds{\color{#f00}{\pi^{3} \over 45} = \color{#f00}{{2 \over \pi}\sum_{n = 1}^{\infty}{1 \over n^{4}}}\quad \implies\bbx{\ds{\sum_{n = 1}^{\infty}{1 \over n^{4}} = {\pi^{4} \over 90}}}}$