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. 
 A: 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:
$$ \int_{0}^{\infty} \bigg( \sum_{n=1}^{\infty}\frac{x^n }{n!} \bigg)\frac{e^{-2x}}{1 - e^{-x}} \, dx
= \int_{0}^{\infty} e^{-x} \, dx
< \infty. $$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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.

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