# Proving that $\left(\frac{\pi}{2}\right)^{2}=1+\sum_{k=1}^{\infty}\frac{(2k-1)\zeta(2k)}{2^{2k-1}}$.

Wolfram$\alpha$ says that we have the following identity $$\left(\frac{\pi}{2}\right)^{2}=1+2\sum_{k=1}^{\infty}\frac{(2k-1)\zeta(2k)}{2^{2k}}$$

but, how does one prove such identity?

• Maybe $$\zeta(2k)=(-1)^{k+1}\frac{B_{2k}(2\pi)^{2k}}{2(2k)!}$$ might help? The Bernoulli numbers might be troublesome though. Dec 18, 2013 at 16:27
• How about utilizing the Taylor series of the log-gamma function? Dec 18, 2013 at 16:35

It's well known that $$\pi\cot(\pi x)=\frac 1x-2\sum_{n=1}^\infty\zeta(2n)x^{2n-1};$$ taking derivatives we get $$\frac{\pi^2}{\sin^2\pi x}=x^{-2}+2\sum_{n=1}^\infty(2n-1)\zeta(2n)x^{2n-2};$$ in particular for $x=\frac12$ $$\pi^2=4+2\sum_{n=1}^\infty(2n-1)\zeta(2n)\frac1{2^{2n-2}};$$ this is your formula (multiplied by 4).

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$\pars{\pi \over 2}^{2} = 1 + 2\sum_{k=1}^{\infty}{\pars{2k - 1}\zeta\pars{2k} \over 2^{2k}}:\ {\Large ?} \tag{1}$$

\begin{align} &\color{#00f}{\large\sum_{k=1}^{\infty}{\pars{2k - 1}\zeta\pars{2k} \over 2^{2k}}}= \sum_{k=1}^{\infty}{2k - 1 \over 2^{2k}}\sum_{\ell = 1}^{\infty}{1 \over \ell^{2k}} =\sum_{\ell = 1}^{\infty}\sum_{k=1}^{\infty}\pars{2k - 1}\pars{1 \over 2\ell}^{2k} \\[3mm]&=\sum_{\ell = 1}^{\infty}\bracks{x^{2} \totald{}{x}\sum_{k=1}^{\infty}x^{2k - 1}}_{x = 1/\pars{2\ell}} =\sum_{\ell = 1}^{\infty}\braces{x^{2}\, \totald{}{x}\bracks{x \over 1 - x^{2}}}_{x = 1/\pars{2\ell}} \\[3mm]&=\sum_{\ell = 1}^{\infty}\bracks{x^{2}\, {1 + x^{2} \over \pars{1 - x^{2}}^{2}}}_{x = 1/\pars{2\ell}} =\sum_{\ell = 1}^{\infty}\bracks{ {1/x^{2} + 1 \over \pars{1/x^{2} - 1}^{2}}}_{x = 1/\pars{2\ell}} \\[3mm]&=\sum_{\ell = 1}^{\infty}\bracks{ {1 \over 1/x^{2} - 1} + {2 \over \pars{1/x^{2} - 1}^{2}}}_{x = 1/\pars{2\ell}} =\left.\pars{2\,\totald{}{\mu} + 1}\sum_{\ell = 1}^{\infty}{1 \over 4\ell^{2} - \mu} \right\vert_{\mu = 1} \\[3mm]&=\pars{2\,\totald{}{\mu} + 1}\bracks{% {2 - \root{\mu}\pi\cot\pars{\root{\mu}\pi/2}} \over 4\mu}_{\mu = 1} =\color{#00f}{\large\half\bracks{\pars{\pi \over 2}^{2} - 1}}\tag{2} \end{align}

Replace $\pars{2}$ in $\pars{1}$.

Do not worry about the downvote. It is just an attack on my answers.

The series can have the following integral representation

$$\sum_{k=1}^{\infty}\frac{(2k-1)\zeta(2k)}{2^{2k-1}} = \int_{0}^{1}\frac{(t^2+1)\ln t}{t^2-1} dt = \frac{\pi^2}{4}-1.$$

• Interesting. But why first equality holds? Dec 18, 2013 at 17:10
• @GrigoryM: The series admits this integral representation. Dec 18, 2013 at 17:15
• @Downvoter: What's the downvote for? Dec 26, 2013 at 21:08
• @Downvoter: What's the downvote for? You do not like this integral representation! Tell us why you down voted the answer. Dec 29, 2013 at 1:01
• None of downvotes are mine, but how exactly 'this series admits this integral representation — prove it yourself' is any better than 'this series is equal to $\pi^2/4-1$ — prove it yourself'? Maybe experts can see what you're alluding too, but this answer hasn't taught me anything, I'm afraid (and, AFAICS, other users haven't found this helpful either). Jan 25, 2014 at 22:26