# 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.

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

• @FelixMarin Yes, I'm an idiot :D – Marco Cantarini Nov 19 '16 at 8:51
• It happens sometimes: we says $2 + 2 =5$ several times. After sleeping, we recover the $4$. You're not an idiot. You are very clever. – Felix Marin Nov 19 '16 at 8:53

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:

$$\int_{0}^{\infty} \bigg( \sum_{n=1}^{\infty}\frac{x^n }{n!} \bigg)\frac{e^{-2x}}{1 - e^{-x}} \, dx = \int_{0}^{\infty} e^{-x} \, dx < \infty.$$

• "The computation is not hard": haha. – Yves Daoust Nov 18 '16 at 16:59

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