Vtyjkyuk

For some time I've been playing with this kind of sums, for example I was able to find that
$$
fracpi2=1+2sum_k=1^inftyleft( zeta(2k+1)-beta(2k+1)right)
$$
where
$$
beta(x)=sum_k=1^inftyfrac(-1)^k-1(2k-1)^x
$$
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
$$
beta(2k+1)=(-1)^kfracE_2kpi^2k+14^k+1(2k)!.
$$

Where $E_2k$ are the Euler's numbers:
$$
beginmatrix
E_0 &=& 1\
E_2 &=& -1\
E_4 &=& 5\
E_6 &=& -61\
E_8 &=& 1385\
vdots &=&vdots
endmatrix
$$
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
$$
sum_k=1^inftyleft( zeta(2k)-beta(2k)right)
$$
Any help would be appreciated.

Thanks.

There is a closed form for your series:

$$
sum_k=1^inftyleft( zeta(2k)-beta(2k)right)=frac12+frac12ln2. tag1
$$

Proof.
Using absolute convergence of the series, you may write

$$beginalign
sum_k=1^inftyleft( zeta(2k)-beta(2k)right) & = sum_k=1^inftyleft( sum_n=1^inftyfrac1n^2k-sum_n=1^inftyfrac(-1)^n-1(2n-1)^2kright)\\
& = sum_k=1^inftyleft( sum_n=colorred2^inftyfrac1n^2k-sum_n=colorred2^inftyfrac(-1)^n-1(2n-1)^2kright)\\
& = sum_n=2^inftyleft( sum_k=1^inftyfrac1n^2k-sum_k=1^inftyfrac(-1)^n-1(2n-1)^2kright)\\
& = sum_n=2^inftyleft( frac1n^2frac11-frac1n^2+frac(-1)^n(2n-1)^2frac11-frac1(2n-1)^2right)\\
& = sum_n=2^inftyfrac1(n+1)(n-1)+sum_n=2^inftyfrac18n(2n-1)-sum_n=2^inftyfrac18(2n-1)(n-1)\\
& = frac12+frac12ln2,
endalign$$ where, in the last steps, we have used partial fraction decomposition, telescoping terms and a familiar series for $ln 2$.

Hint: Rewrite the general term as an infinite series, and then switch the order of summation. :-$)$

$$
beginalign
colorredS,&=sum_n=1^inftyleft[,zeta(2n)-beta(2n),right] =sum_n=1^inftyfracGamma(2n)zeta(2n)-Gamma(2n)beta(2n)(2n-1)! \[2mm]
&=int_0^inftyleft(frac1e^x-1-frace^xe^2x+1right)left(sum_n=1^inftyfracx^2n-1(2n-1)!right),dxqquadcolorbluesmallleftsinh(x)=sum_n=1^inftyfracx^2n-1(2n-1)!right \[2mm]
&=int_0^inftyleft(frace^x+1e^2x-1-frace^xe^2x+1right)left(frace^2x-12e^xright),dx =frac12int_0^inftyleft(1+frac1e^x-frace^2x-1e^2x+1right),dx \[2mm]
&=frac12int_0^inftyleft(frac1e^x+frac2e^2x+1right),dx =colorredfrac12left[,Gamma(1)+eta(1),right] =frac12left(1+log(2)right)
endalign
$$