calculus - Compute $sumlimits_{n = 1}^{infty} frac{1}{n^4}$.

Compute the Fourier series for $x^3$ and use it to compute the value of $\sum\limits_{n = 1}^{\infty} \frac{1}{n^4}$.

I determined the coefficients of the Fourier series, which are

$$a_0 = \dfrac{\pi^3}{2}; \qquad a_n = \dfrac{6(\pi^2 n^2 - 2)(-1)^n + 12}{\pi n^4}$$

Then, I get

$$x^3 = \dfrac{\pi^3}{4} + \sum\limits_{n = 1}^{\infty} \dfrac{6(\pi^2 n^2 - 2)(-1)^n + 12}{\pi n^4}\cos(nx)$$

If $x = \pi$, then

$$\begin{aligned} \pi^3 &= \dfrac{\pi^3}{4} + \sum\limits_{n = 1}^{\infty} \dfrac{6(\pi^2 n^2 - 2)(-1)^n + 12}{\pi n^4}\cos(n\pi)\\ \dfrac{3\pi^3}{4} &= \sum\limits_{n = 1}^{\infty} \dfrac{6(\pi^2 n^2 - 2)(-1)^n + 12}{\pi n^4}(-1)^n \end{aligned}$$

I'm stuck. It's easy to compute $\sum\limits_{n = 1}^{\infty} \frac{1}{n^2}$, using the Fourier series, but for this type of problem I'm stuck.

Any comments or suggestions? By the way, I know that

$$\sum\limits_{n = 1}^{\infty} \dfrac{1}{n^4} = \dfrac{\pi^4}{90}$$

I need to know how to get there.

Answer

From your identity

$$\begin{equation*} \frac{3\pi ^{3}}{4}=\sum_{n=1}^{\infty }\frac{6(\pi ^{2}n^{2}-2)(-1)^{n}+12}{ \pi n^{4}}(-1)^{n} \end{equation*}$$

expanding the right hand side and using the result $\sum_{n=1}^{\infty }\frac{1}{n^{2}}=\frac{\pi^2}{6}$, we get

$$\begin{eqnarray*} \frac{3\pi ^{4}}{4} &=&\sum_{n=1}^{\infty }\frac{6(\pi ^{2}n^{2}-2)(-1)^{n}+12}{n^{4}}(-1)^{n} \\ &=&6\pi ^{2}\sum_{n=1}^{\infty }\frac{1}{n^{2}}-12\sum_{n=1}^{\infty }\frac{1 }{n^{4}}+12\sum_{n=1}^{\infty }\frac{(-1)^{n}}{n^{4}} \\ &=&\pi ^{4}-12\sum_{n=1}^{\infty }\frac{1}{n^{4}}-12\sum_{n=1}^{\infty } \frac{(-1)^{n-1}}{n^{4}}. \end{eqnarray*}$$

Now we need to express the alternating series $\sum_{n=1}^{\infty }\frac{ (-1)^{n-1}}{n^{4}}$ in terms of $\sum_{n=1}^{\infty }\frac{1}{n^{4}}$, e.g.
as follows

$$\begin{eqnarray*} \sum_{n=1}^{\infty }\frac{\left( -1\right) ^{n-1}}{n^{4}} &=&\sum_{n=1}^{ \infty }\frac{1}{n^{4}}-2\sum_{n=1}^{\infty }\frac{1}{(2n)^{4}} =\sum_{n=1}^{\infty }\frac{1}{n^{4}}-\frac{1}{2^{3}}\sum_{n=1}^{\infty } \frac{1}{n^{4}}=\frac{7}{8}\sum_{n=1}^{\infty }\frac{1}{n^{4}}. \end{eqnarray*}$$

Then

$$\begin{eqnarray*} \frac{3\pi ^{4}}{4} &=&\pi ^{4}-12\sum_{n=1}^{\infty }\frac{1}{n^{4}}-\frac{ 21}{2}\sum_{n=1}^{\infty }\frac{1}{n^{4}} \\ &=&\pi ^{4}-\frac{45}{2}\sum_{n=1}^{\infty }\frac{1}{n^{4}}. \end{eqnarray*}$$

Solving for $\sum_{n=1}^{\infty }\frac{1}{n^{4}}$ we finally obtain

$$\begin{equation*} \sum_{n=1}^{\infty }\frac{1}{n^{4}}=\frac{2}{45}\left( \pi ^{4}-\frac{3\pi ^{4}}{4}\right) =\frac{\pi ^{4}}{90}. \end{equation*}$$