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. 

The computation is not hard:
\begin{align*}
\sum_{n=1}^{\infty}[\zeta(2n) - \zeta(2n+1)]
&= \sum_{n=1}^{\infty}(-1)^{n+1}[\zeta(n+1) - 1] \\
&= \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n!}\sum_{k=2}^{\infty} \frac{n!}{k^{n+1}} \\
&= \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n!}\sum_{k=2}^{\infty} \int_{0}^{\infty} x^n e^{-kx} \, dx \\
&\stackrel{(1)}{=} \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n!}\int_{0}^{\infty} \frac{x^n e^{-2x}}{1 - e^{-x}} \, dx \\
&\stackrel{(2)}{=} \int_{0}^{\infty} \bigg( \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n!} x^n  \bigg)\frac{e^{-2x}}{1 - e^{-x}} \, dx \\
&= \int_{0}^{\infty} e^{-2x} \, dx
 = \frac{1}{2}.
\end{align*}
For $\text{(1)}$, we utilized Tonelli's theorem to interchange the order of integral and summation. For $\text{(2)}$, we utilized Fubini's theorem along with the estimate:
$$ \int_{0}^{\infty} \bigg( \sum_{n=1}^{\infty}\frac{x^n }{n!} \bigg)\frac{e^{-2x}}{1 - e^{-x}} \, dx
= \int_{0}^{\infty} e^{-x} \, dx
< \infty. $$
