Looking for a closed form for $\sum_{k=1}^{\infty}\left( \zeta(2k)-\beta(2k)\right)$ 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.
 A: Hint: Rewrite the general term as an infinite series, and then switch the order of summation. :-$)$
A: There is a closed form for your series:

$$
\sum_{k=1}^{\infty}\left( \zeta(2k)-\beta(2k)\right)=\frac{1}{2}+\frac{1}{2}\ln2. \tag1
$$

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 $\ln 2$.
A: $$ 
\begin{align} 
\color{red}{S}\,&=\sum_{n=1}^{\infty}\left[\,\zeta(2n)-\beta(2n)\,\right] =\sum_{n=1}^{\infty}\frac{\Gamma(2n)\zeta(2n)-\Gamma(2n)\beta(2n)}{(2n-1)!} \\[2mm] 
&=\int_{0}^{\infty}\left(\frac{1}{e^{x}-1}-\frac{e^{x}}{e^{2x}+1}\right)\left(\sum_{n=1}^{\infty}\frac{x^{2n-1}}{(2n-1)!}\right)\,dx\qquad\color{blue}{\small\left\{\sinh(x)=\sum_{n=1}^{\infty}\frac{x^{2n-1}}{(2n-1)!}\right\}} \\[2mm] 
&=\int_{0}^{\infty}\left(\frac{e^x+1}{e^{2x}-1}-\frac{e^{x}}{e^{2x}+1}\right)\left(\frac{e^{2x}-1}{2e^{x}}\right)\,dx =\frac{1}{2}\int_{0}^{\infty}\left(1+\frac{1}{e^{x}}-\frac{e^{2x}-1}{e^{2x}+1}\right)\,dx \\[2mm] 
&=\frac{1}{2}\int_{0}^{\infty}\left(\frac{1}{e^{x}}+\frac{2}{e^{2x}+1}\right)\,dx =\color{red}{\frac{1}{2}\left[\,\Gamma(1)+\eta(1)\,\right]} =\frac{1}{2}\left(1+\log(2)\right) 
\end{align} 
$$ 
