how do I prove that $$\frac{ζ（2n）}{Π^{2n}}-\frac{ζ(2n-2)}{3!Π^{2n-2}}+\frac{ζ（2n-4）}{5！Π^{2n-4}}-……+\frac{(-1)^{n-1}ζ(2)}{(2n-1)!Π^2}+\frac{(-1)^nn}{(2n+1)!} = 0$$ maybe i can use $\sum_{n=1}^∞ \frac{z^{2n-1}}{(2n-1)!}{(-1)^{n-1}} = sinz$
but i have no idea and really need your help
Thank you!
You need to use $$\frac{\pi^2}{\sin^2 (\pi z)} = \sum_n \frac1{(z+n)^2}= z^{-2}+\sum_{k\ge 1} 2\zeta(2k)z^{2k-2} $$ Then multiply both sides by $\sin^2(\pi z)$'s Taylor series