I have: $$g(x)=x(1-|x|), \space\space\space\space\space\space -1\leq x \lt 1$$
From which I got the Fourier series: $$Sg(t)=\frac{-8}{\pi^3}\sum_{n=0}^{\infty}\frac{1}{(2n+1)^3}\sin((2n+1)x)$$
The next assignment is to calculate the following sums using the Fourier series:
$$\sum_{n=0}^{\infty}\frac{1}{(2n+1)^6}$$ and
$$\sum_{n=0}^{\infty}\frac{-1^n}{(2n+1)^3}$$
I managed to get the first one using Parsevals formula:
$$\frac{1}{2}\sum_{n=0}^{\infty}(\frac{-8}{\pi^3}\frac{1}{(2n+1)^3})^2=\frac{1}{2}\int_{-1}^1(x(1-|x|))^2dx$$$$\Rightarrow$$
$$\frac{32}{\pi^6}\sum_{n=0}^{\infty}\frac{1}{(2n+1)^6}=\frac{1}{30}$$$$\Rightarrow$$
$$\sum_{n=0}^{\infty}\frac{1}{(2n+1)^6}=\frac{\pi^6}{960}$$
My question is how to go about the second one.
Thank you in advance for the help.
Best Answer
The Fourier Series of $g$ converges to $g$ uniformly thus pointwise.
For $x=\frac{\pi}{2}$ you have that $\sin{((2n+1)\frac{\pi}{2})}=(-1)^n$