Obtain sum of a sequence from sum of its odd terms.

Question

I would like to compute the sum $$ \sum_{k=1}^\infty \frac{1}{k^4} $$ by using the Fourier series of $f(x)=|x|$ over $(-\pi,\pi)$. Coefficients $b_k$ are all $0$ because $f$ is even. Doing the integration stuff, I obtained: $$ a_0 = \pi $$ and $$ a_k = \frac{2}{k^2}\bigg((-1)^k-1\bigg) $$ for $k>0$. The Parseval's equality gives: $$ \frac{a_0^2}{2} + \sum_{k=1}^\infty (a_k^2+b_k^2)= \frac{1}{\pi}\int_{-\pi}^{\pi}f^2dx $$ which gives $$ \frac{\pi^2}{2} + \sum_{k=1}^\infty \frac{4}{\pi^2k^4}(2-2(-1)^k) = \frac{2}{3}\pi^2 $$ which simplifies to $$ \sum_{k=1}^\infty \frac{1}{k^4} - \sum_{k=1}^\infty \frac{(-1)^k}{k^4} = \frac{\pi^4}{48} $$ which basically says: $$ \sum_{k=0}^\infty \frac{1}{(2k+1)^4}=\frac{\pi^4}{96} $$ any idea how to obtain the sum from there?

Answer 1

Observe that what you have is that $2\sum_{k=0}^{\infty} \frac 1{(2k+1)^4}=\frac {\pi^4}{48}$. Calling $\sum_{k=0}^{\infty} \frac 1{k^4}=S$ you have that $\sum_{k=0}^{\infty} \frac 1{(2k)^4}=\frac 1{16} S$ and finally you have $S-\frac 1{16}S=\frac 12 \frac {\pi^4}{48}$ from which $S=\frac {\pi^4}{90}$

Answer 2

You essentially have

$${\frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + ... = \frac{\pi^4}{96}}$$

You want to find

$${\frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + ... = ?}$$

in other words, you want to add on

$${\frac{1}{2^4} + \frac{1}{4^4} + ...}$$

Factoring out a ${\frac{1}{2^4}}$ on the above yields

$${\frac{1}{2^4}\left(\frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + ...\right)}$$

So overall, if you call ${S=\frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + ...}$ you have

$${\left(\frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + ...\right) + \left(\frac{1}{2^4} + \frac{1}{4^4} + ...\right) = S}$$

$${\Rightarrow \frac{\pi^4}{96} + \frac{1}{2^4}S = S}$$

Can you now rearrange for ${S}$?