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$$ \sum_{k=0}^\infty (\zeta(2k+3) -1) = \frac{1}{4} $$ I found this nice identity and want to prove it, but I don't know how. My first thought was that I can create a geometric series with the same limit like: $$\sum_{k=0}^\infty \frac{1}{8} \cdot \frac{1}{2^k} = \frac{1}{4}$$ But I have no idea, how to write the first one in a way, that these are the same. They are obviously only equal, if we sum infinitely many terms together. Do you have ideas, how to proof this identity or do you know, where the proof can be found? Thank you for your answers!

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    $\begingroup$ Where did you found it? All terms are greater than 1! $\endgroup$ – gammatester Mar 30 '18 at 13:41
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    $\begingroup$ You have a typo, should be $\sum_{k=0}^\infty (\zeta(2k+3) - \color{red}{1}) = \frac14$. $\endgroup$ – achille hui Mar 30 '18 at 13:50
  • $\begingroup$ Actually , this sum seems to be $k+\frac{5}{4}$ with a very small error already for $k=10$ $\endgroup$ – Peter Mar 30 '18 at 13:51
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There is a typo in the question. The sum $\sum\limits_{k=0}^\infty \zeta(2k+3)$ diverges because $\lim\limits_{k\to\infty} \zeta(2k+3) = 1$.
The correct statement should be

$$\sum_{k=0}^\infty (\zeta(2k+3) - 1) = \frac14$$

The proof is relatively simple,

$$\begin{align} \sum_{k=0}^\infty (\zeta(2k+3)-1) = &\sum_{k=0}^\infty \sum_{n=2}^\infty \frac{1}{n^{2k+3}} \stackrel{\color{blue}{[1]}}{=} \sum_{n=2}^\infty \sum_{k=0}^\infty \frac{1}{n^{2k+3}} = \sum_{n=2}^\infty \frac{1}{n^3}\sum_{k=0}^\infty (n^{-2})^k\\ = &\sum_{n=2}^\infty \frac{1}{n^3}\frac{1}{1 - n^{-2}} = \sum_{n=2}^\infty \frac{1}{(n-1)n(n+1)} = \sum_{n=2}^\infty \frac12\left(\frac{1}{(n-1)n} - \frac{1}{n(n+1)}\right)\\ \stackrel{\color{blue}{[2]}}{=} & \frac12\left(\frac{1}{(2-1)2}\right) = \frac14\end{align} $$

Notes

  • $\color{blue}{[1]}$ - all terms are non-negative, it is legal to exchange order of summation.

  • $\color{blue}{[2]}$ - the sum is a telescoping one.

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  • $\begingroup$ Wow, so easy is that (if one has the "know-how") (+1) $\endgroup$ – Peter Mar 30 '18 at 14:16
  • $\begingroup$ The concept of a telescoping sum is very fascinating. I was not aware of this type of series $\endgroup$ – Mister Set Mar 30 '18 at 17:15
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By the integral representation of $\zeta\left(s\right)$ we have $$\zeta\left(2k+3\right)-1=\frac{1}{\left(2k+2\right)!}\int_{0}^{\infty}\frac{u^{2k+2}}{e^{u}\left(e^{u}-1\right)}du$$ then $$\sum_{k\geq0}\left(\zeta\left(2k+3\right)-1\right)=\int_{0}^{\infty}\frac{1}{e^{u}\left(e^{u}-1\right)}\sum_{k\geq0}\frac{u^{2k+2}}{\left(2k+2\right)!}du$$ $$=\int_{0}^{\infty}\frac{\cosh\left(u\right)-1}{e^{u}\left(e^{u}-1\right)}du=\frac{1}{2}\int_{0}^{\infty}e^{-u}-e^{-2u}du=\color {red}{\frac{1}{4}}.$$

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\ds{\sum_{k = 0}^{\infty}\bracks{\zeta\pars{2k + 3} - 1}}} = \sum_{k = 1}^{\infty}\bracks{\zeta\pars{2k + 1} - 1} \\[5mm] = &\ \lim_{z \to 1}\bracks{{1 \over 2z} - {1 \over 2}\,\pi\,\cot\pars{\pi z} - {1 \over 1 - z^{2}} + 1 - \gamma - \Psi\pars{1 + z}} = \bbx{1 \over 4} \end{align}

where $\ds{\Psi}$ is the Digamma Function. See $\ds{\mathbf{\color{black}{6.3.15}}}$ in A & S Table.

Note that $\ds{\Psi\pars{2} = \Psi\pars{1} + 1 = -\gamma + 1}$.

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