I've been given the half-range sine series
$$f(\pi-t)=\sum_{n=1}^\infty \frac{\sin(2n-1)t}{(2n-1)^3}, \ \ \ \ \ \ \ 0\le t\le \pi$$
The value of series $\sum_{n=1}^\infty \frac{1}{(2n-1)^6}$ needs to be determined.
Since $f(x)$ is not directly known, how are we supposed to apply Parseval's Theorem. Any help is appreciated.
Using the definition of the zeta function, you have $$\sum_{i=1}^\infty\frac{1}{(2i-1)^n}=(1-2^{-n})\,\zeta (n)$$ and the nice values for even $n$ $$\left( \begin{array}{cc} n & \zeta (n) \\ 2 & \frac{\pi ^2}{6} \\ 4 & \frac{\pi ^4}{90} \\ 6 & \frac{\pi ^6}{945} \\ 8 & \frac{\pi ^8}{9450} \\ 10 & \frac{\pi ^{10}}{93555} \end{array} \right)$$