Obtain sum of a sequence from sum of its odd terms. 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?
 A: 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 $S=\frac {\pi^4}{90}$
A: 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 ${S}$?
