Use Parseval's identity for series with $\sin((2k+1)x)$.

Question

I'm trying to use Parseval's identity to evaluate the values of the series $$\sum_{k=0}^{\infty}\frac{1}{(2k+1)^6}$$ using a Fourier series that I have derived earlier as $f(x)=x(\pi-|x|) = \sum_{k=0}^\infty \frac{8}{\pi (2k+1)^3}\sin((2k+1)x)$. What I know is a special case of the Parseval identity, namely:$${\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x)^{2}\,\mathrm {d} x={\frac {a_{0}^{2}}{2}}+\sum _{k=1}^{\infty }(a_{k}^{2}+b_{k}^{2}).$$ I was able to use this to evaluate $\sum_{n=1}^\infty \frac{1}{n^4}$ given $g(x)=x^2=\frac{pi^2}{3}+\sum_{n=1}^{\infty}\frac{4(-1)^n}{n^2}\cos(nx)$ which was straightforward but I'm struggling with $f(x)$ since the series has a different form. Am I supposed to express the part with $\sin((2k+1)x)$ in a different way so that I can apply the Parseval identity?

Answer 1

But if you know the sum of $\;1/n^6\;$ then you're done:

$$\frac{\pi^6}{945}=\sum_{n=1}^\infty\frac1{n^6}=\sum_{n=1}^\infty\frac1{(2n)^6}+\sum_{n=1}^\infty\frac1{(2n-1)^6}=\frac1{64}\frac{\pi^6}{945}+\sum_{n=1}^\infty\frac1{(2n-1)^6}\implies$$

$$\sum_{n=1}^\infty\frac1{(2n-1)^6}=\left(1-\frac1{64}\right)\frac{\pi^6}{945}=\ldots$$