convergence divergence - Is this :$sum_{n=1}^{infty }zeta(2n)-zeta(2n+1)=frac{1}{2}$?

I would like to know the behavior of the Riemann zeta function values at even
and odd integers for studying irrationality between those values. I have tried using wolfram alpha to check the value of this sum:
$$
\sum_{n = 1}^{\infty}\left[\zeta\left(2n\right)-\zeta\left(2n + 1\right)\right].

$$
It tells me it equals $\frac{1}{2}$ .

Note: I don't have any method to show if the titled sum is true . Maybe I find who is help me here for evaluating the titled sum.

Thanks for any help.

Answer

Note that , due to the absolute convergence, we have$$\sum_{n\geq1}\left(\zeta\left(2n\right)-\zeta\left(2n+1\right)\right)=\sum_{n\geq1}\left(\sum_{k\geq1}\frac{1}{k^{2n}}-\sum_{k\geq1}\frac{1}{k^{2n+1}}\right)
$$ $$=\sum_{n\geq1}\sum_{k\geq1}\frac{k-1}{k^{2n+1}}=\sum_{k\geq2}\sum_{n\geq1}\frac{k-1}{k^{2n+1}}
$$ $$=\sum_{k\geq2}\frac{1}{k\left(k+1\right)}=\sum_{k\geq2}\left(\frac{1}{k}-\frac{1}{k+1}\right)=\color{red}{\frac{1}{2}}.$$