I was playing around recently with even integer inputs for the Rriemann Zeta Function, calculating their values via the Fourier series for $x^2,x^4,x^6...$ from $[-\pi;\pi]$ then plugging in $\pi$ and rearranging to isolate the $\zeta$ term. For example, if we wish to find $\zeta(2)$, we find the fourier series for $x^2$ on $[-\pi;\pi]$: $$x^2=\frac{\pi^2}{3}+\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\cos(nx)$$ Substitutung $\pi$ and rearranging yields: $$\zeta(2)=\frac{1}{4}(\pi^2-\frac{\pi^2}{3})$$ Going through the same process to find $\zeta(4)$, we find that it's value depends on $\zeta(2)$: $$\zeta(4)=\frac{1}{-48}(\pi^4-\frac{\pi^4}{5}-8\pi^2\zeta(2))$$ Or, for $\zeta(6)$: $$\zeta(6)=\frac{1}{1440}(\pi^6-\frac{\pi^6}{7}-12\pi^4\zeta(2)+240\pi^2\zeta(4))$$ So it was only natural for me to conject that: $$\zeta(2s)=\frac{1}{2\left(2s\right)!\left(-1\right)^{s+1}}\left(\pi^{2s}\left(1-\frac{1}{2s+1}\right)+2\sum_{n=1}^{s-1}\frac{\left(-1\right)^{n}\left(2s\right)!}{\left(2s-2n+1\right)!}\pi^{2s-2n}\zeta\left(2n\right)\right)$$ Or, to re-arrange terms slightly: $$\zeta(2s)=-\pi^{2s}\left(-1\right)^{s}\left(\frac{s}{\left(2s+1\right)!}+\sum_{n=1}^{s-1}\frac{\left(-1\right)^{n}}{\left(2s-2n+1\right)!}\pi^{-2n}\zeta\left(2n\right)\right)$$ $\forall s\in\mathbb{N}^*$ Where for $s=1$ the sum doesn't exist. I know this sum checks out for at least the first ten terms but I would ideally like to find a way to prove my conjecture. Is it already well known, can it be proven or disproven? This is purely recreational but I would like to have some sort of closure on this problem.
Thank you for your time!
Put $a_n=\zeta(2n)/\pi^{2n}$, then for $|z|<\pi$, using the infinite product for $\sin z$, we have $$\sum_{n>0}a_n z^{2n}=\sum_{n,k>0}\left(\frac{z}{k\pi}\right)^{2n}=\sum_{k>0}\frac{z^2}{k^2\pi^2-z^2} =\frac{z}{2}\frac{d}{dz}\log\frac{z}{\sin z}=\frac{1-z\cot z}2.$$
Multiplying this by $\sin z$ (as its power series this time), we obtain $$\left(\sum_{n>0}a_n z^{2n}\right)\left(\sum_{k\geqslant 0}\frac{(-1)^k z^{2k+1}}{(2k+1)!}\right)=\frac{\sin z-z\cos z}{2}=\sum_{n>0}\frac{(-1)^{n-1}n}{(2n+1)!}z^{2n+1}$$ or, extracting the coefficient of $z^{2n+1}$, $$\sum_{k=1}^n\frac{(-1)^{k-1}a_k}{(2n-2k+1)!}=\frac{n}{(2n+1)!}.$$
This is equivalent to the equality in question.