For some time I've been playing with this kind of sums, for example I was able to find that
$$
\frac{\pi}{2}=1+2\sum_{k=1}^{\infty}\left( \zeta(2k+1)-\beta(2k+1)\right)
$$
where
$$
\beta(x)=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{(2k-1)^{x}}
$$
is the Dirichlet's beta function and $\zeta(x)$ is the Riemann's zeta function. I find this result very interesting, because we know that for odd integers $\beta(x)$ reduces to
$$
\beta(2k+1)=(-1)^{k}\frac{E_{2k}\pi^{2k+1}}{4^{k+1(2k)!}}.
$$
Where $E_{2k}$ are the Euler's numbers:
$$
\begin{matrix}
E_{0} &=& 1\\
E_{2} &=& -1\\
E_{4} &=& 5\\
E_{6} &=& -61\\
E_{8} &=& 1385\\
\vdots &=&\vdots
\end{matrix}
$$
But there is no known similar simple relation for $\zeta(2k+1)$. Nevertheless, when both are combined they give the above beautiful result.
Now, I'd like to know if there is something similar for $$ \sum_{k=1}^{\infty}\left( \zeta(2k)-\beta(2k)\right) $$ Any help would be appreciated.
Thanks.
There is a closed form for your series:
Proof. Using absolute convergence of the series, you may write
$$\begin{align} \sum_{k=1}^{\infty}\left( \zeta(2k)-\beta(2k)\right) & = \sum_{k=1}^{\infty}\left( \sum_{n=1}^{\infty}\frac{1}{n^{2k}}-\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(2n-1)^{2k}}\right)\\\\ & = \sum_{k=1}^{\infty}\left( \sum_{n=\color{red}2}^{\infty}\frac{1}{n^{2k}}-\sum_{n=\color{red}2}^{\infty}\frac{(-1)^{n-1}}{(2n-1)^{2k}}\right)\\\\ & = \sum_{n=2}^{\infty}\left( \sum_{k=1}^{\infty}\frac{1}{n^{2k}}-\sum_{k=1}^{\infty}\frac{(-1)^{n-1}}{(2n-1)^{2k}}\right)\\\\ & = \sum_{n=2}^{\infty}\left( \frac{1}{n^{2}}\frac{1}{1-\frac{1}{n^{2}}}+\frac{(-1)^{n}}{(2n-1)^{2}}\frac{1}{1-\frac{1}{(2n-1)^{2}}}\right)\\\\ & = \sum_{n=2}^{\infty}\frac{1}{(n+1)(n-1)}+\sum_{n=2}^{\infty}\frac{1}{8n(2n-1)}-\sum_{n=2}^{\infty}\frac{1}{8(2n-1)(n-1)}\\\\ & = \frac{1}{2}+\frac{1}{2}\ln2, \end{align}$$ where, in the last steps, we have used partial fraction decomposition, telescoping terms and a familiar series for $\ln 2$.