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. 

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\sum_{n = 1}^{\infty}\bracks{\zeta\pars{2n}-\zeta\pars{2n + 1}} =
\overbrace{%
\sum_{n = 1}^{\infty}\bracks{\zeta\pars{2n} - 1}}^{\ds{3 \over 4}}\ -\
\overbrace{%
\sum_{n = 1}^{\infty}\bracks{\zeta\pars{2n + 1} - 1}}^{\ds{1 \over 4}} =
\bbx{1 \over 2}
\end{align}

The 'even' and 'odd' above sums are well known identities.

