If $n$ is a natural number and $S_n=\sum_{k=1}^{\infty }\frac{1}{(4k^2-1)^n}$,
then $S_1=\frac{1}{2}, S_2=\frac{\pi^2-8}{16}, S_3=\frac{-3\pi^2+32}{64}, S_4=\frac{\pi^4+30\pi^2-384}{768},\dots$.
How $S_1, S_2, S_3, S_4$ are evaluated in exact forms?
How to evaluate $S_n$ for higher values of $n$, say $n=5$?
Best Answer
$$s_1=\frac{1}{2}\sum_{k=1}^{+\infty}\left(\frac{1}{2k-1}-\frac{1}{2k+1}\right)=\frac{1}{2}\lim_{n\rightarrow+\infty}\sum_{k=1}^n\left(\frac{1}{2k-1}-\frac{1}{2k+1}\right)=$$ $$=\frac{1}{2}\lim_{n\rightarrow+\infty}\left(1-\frac{1}{2n+1}\right)=\frac{1}{2}.$$ $$s_2=\frac{1}{4}\sum_{k=1}^{+\infty}\left(\frac{1}{(2k-1)^2}+\frac{1}{(2k+1)^2}-\frac{2}{(2k-1)(2k+1)}\right)=$$ $$=\frac{1}{4}\left(\frac{\pi^2}{8}+\frac{\pi^2}{8}-1-1\right)=\frac{\pi^2-8}{16}.$$ For the rest we can use the similar way.