Consider
$$\sum_{n=1}^{\infty}(-1)^n{1-2^{2n}\over 2^{2n}}\cdot{\zeta(4n+2)\over 2n+1}=S_1\tag1$$ $$\sum_{n=1}^{\infty}\tan^{-1}\left({1\over 4n^6+3n^2}\right)=S_2\tag2$$
How does one determine the closed form for $S_1$ or $S_2$?
We believe that that $S_1=S_2$, but how can we show that?
Through the identity $\zeta(4n+2)=\frac{1}{(4n+1)!}\int_{0}^{+\infty}\frac{x^{4n+1}}{e^x-1}\,dx $ we have
$$ S_1 = \int_{0}^{+\infty}\frac{2}{x(e^x-1)}\left[2\,\text{ssh}\left(\frac{x}{2}\right)-\text{ssh}\left(\frac{x}{\sqrt{2}}\right)\right]\,dx \tag{1}$$
where $\text{ssh}(z)=\sin(z)\sinh(z)$. $(1)$ can be computed from the residue theorem. About $S_2$,
$$ S_2 = \text{arg}\prod_{n\geq 1}\left(1+\frac{i}{4n^6+3n^2}\right)\tag{2}$$
where the product can be computed from the Weierstrass products of $\sin,\sinh,\cos,\cosh$.
Through $(1)$, $(2)$ and the help of a CAS to prove $S_1=S_2$ should not be difficult.