How to Prove : $ \gamma +\ln\left(\frac{\pi}{4}\right) = \sum_{n=2}^{\infty} \frac{(-1)^{n} \zeta{(n)}}{2^{n-1}n} $

Question

How to Prove :

$$ \gamma +\ln\left(\frac{\pi}{4}\right) = \sum_{n=2}^{\infty} \frac{(-1)^{n} \zeta{(n)}}{2^{n-1}n} $$

I have tried looking at Series definitions of the Polygamma function from which we can obtain $\gamma$ but I'm a little bit lost since the given definitions on Wikipedia are not exactly like this one.

Thank you kindly for your help and time.

Answer 1

The problem is readily reduced to the evaluation of well-known infinite sums and products when approaching in the following way \begin{align*} \sum_{n\ge2}\frac{(-1)^n\zeta(n)}{n2^{n-1}}&=2\sum_{n\ge2}\frac{(-1)^n}{n2^n}\sum_{k\ge1}\frac1{k^n}\\ &=2\sum_{k\ge1}\left[\frac1{2k}-\sum_{n\ge1}\frac{(-1)^{n-1}}n\left(\frac1{2k}\right)^n\right]\\ &=2\sum_{k\ge1}\left[\frac1{2k}-\log\left(1+\frac1{2k}\right)\right] \end{align*} Reorder the partial sums as \begin{align*} 2\sum_{k=1}^m\left[\frac1{2k}-\log\left(1+\frac1{2k}\right)\right]&=\sum_{k=1}^m\frac1k-\sum_{k=1}^m\log\left(\left[\frac{2k+1}{2k}\right]^2\right)\\ &=\sum_{k=1}^m\frac1k+\log\left(\prod_{k=1}^m\left[\frac{2k}{2k+1}\right]^2\right)\\ &=\left[\sum_{k=1}^m\frac1k-\log\left(k+\frac12\right)\right]+\log\left(\frac12\prod_{k=1}^m\left[\frac{(2k)^2}{(2k-1)(2k+1)}\right]\right) \end{align*} Passing the limit $n\to\infty$, using a slight variation on the definition of the Euler-Mascheroni constant combined with the Wallis Product, we obtain $$\lim_{m\to\infty}\left[\sum_{k=1}^m\frac1k-\log\left(k+\frac12\right)\right]+\log\left(\frac12\prod_{k=1}^m\left[\frac{(2k)^2}{(2k-1)(2k+1)}\right]\right)=\gamma+\log\left(\frac\pi4\right)$$ Therefore

$$\therefore~\sum_{n\ge2}\frac{(-1)^n\zeta(n)}{n2^{n-1}}~=~\gamma+\log\left(\frac\pi4\right)$$

Note that your given result is incorrect (I suppose you meant to write $\gamma-\log\left(\frac4\pi\right)$ instead). The result already follows before considering partial sums by using a product representation of the Gamma Function.

Answer 2

Lemma:

Let $f(z)=\sum_{n=2}^{\infty} a_nz^n$ be convergent with radius $>1.$ Then:

$$\sum_{n=2}^{\infty} a_n\zeta(n)=\sum_{k=1}^{\infty} f\left(\frac1k\right)$$

Proof:

$$\begin{align}\sum_{n=2}^{\infty} a_n\zeta(n)&=\sum_{n=2}^{\infty} a_n\sum_{k=1}^{\infty} \frac{1}{k^n} \\ &=\sum_{k=1}^{\infty}\sum_{n=2}^{\infty}a_n\left(\frac 1k\right)^n\\ &=\sum_{k=1}^{\infty}f\left(\frac1k\right) \end{align}$$

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Now, in your case, $a_n=\frac{(-1)^{n}}{2^{n-1}n}$ gives $$f(z)=2\sum_{n=2} \frac{(-z/2)^n}{n}=z-2\log(1+z/2)$$

Now, $$\sum_{k=1}^{N}f(1/k)=H_N - 2\log\left(\frac{3}{2}\cdot \frac{5}{4}\cdots\frac{2N+1}{2N}\right)$$

Now, $H_N-\log N\to \gamma.$ So the limit is equal to the limit $$\gamma -2 \log\left(\frac{3}{2}\cdot \frac{5}{4}\cdots\frac{2N+1}{2N}\cdot\frac{1}{\sqrt{N}}\right)$$ as $N\to\infty.$

Thus, you just need to show:

$$\lim_{N\to\infty}\frac{3}{2}\cdot \frac{5}{4}\cdots\frac{2N+1}{2N}\cdot\frac{1}{\sqrt{N}}=\frac{2}{\sqrt{\pi}}$$

But: $$\frac{3}{2}\cdot \frac{5}{4}\cdots\frac{2N+1}{2N}=\frac{2N+1}{2^{2N}}\binom{2N}{N}$$

And we have that $\binom{2n}{n}\sim \frac{2^{2n}}{\sqrt{\pi n}}$ (see here.)

So we have:

$$\frac{3}{2}\cdot \frac{5}{4}\cdots\frac{2N+1}{2N}\cdot\frac{1}{\sqrt{N}}\sim\frac{2N+1}{N\sqrt{\pi}}\sim \frac{2}{\sqrt{\pi}}$$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{\gamma + \ln\pars{\pi \over 4} = \sum_{n = 2}^{\infty}{\pars{-1}^{n}\,\zeta\pars{n} \over 2^{n - 1}\, n}}:\ {\large ?}}$.

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\begin{align} &\bbox[5px,#ffd]{\sum_{n = 2}^{\infty} {\pars{-1}^{n}\,\zeta\pars{n} \over 2^{n - 1}\, n}} = \sum_{n = 2}^{\infty} {\pars{-1}^{n} \over 2^{n - 1}\, n}\ \overbrace{{1 \over \Gamma\pars{n}}\int_{0}^{\infty}{x^{n - 1} \over \expo{x} - 1}}^{\ds{\zeta\pars{n}}}\ \,\dd x \\[5mm] = &\ 2\int_{0}^{\infty} {\sum_{n = 2}^{\infty}\pars{-x/2}^{n}/n! \over \expo{x} - 1} \,{\dd x \over x} = 2\int_{0}^{\infty} {\expo{-x/2} - 1 + x/2 \over \expo{x} - 1}\,{\dd x \over x} \\[5mm] = &\ \int_{0}^{\infty} {2\expo{-3x/2} - 2\expo{-x} + x\expo{-x} \over 1 - \expo{-x}} \,{\dd x \over x} \\[5mm] \stackrel{x\ =\ -\ln\pars{t}}{=}\,\,\,& -\int_{0}^{1}{2t^{3/2} - 2t -t\ln\pars{t} \over 1 - t} \,{\dd t \over t\ln\pars{t}} \\[5mm] = &\ -\int_{0}^{1}{2t^{3/2} - 2t - t\ln\pars{t} \over 1 - t} \pars{-\int_{0}^{\infty}t^{\xi - 1}\,\dd\xi}\,\dd t \\[5mm] = &\ \int_{0}^{\infty}\int_{0}^{1}{2t^{\xi + 1/2} - 2t^{\xi} - t^{\xi}\ln\pars{t} \over 1 - t}\,\dd t\,\dd\xi \\[5mm] = &\ \int_{0}^{\infty}\bracks{% 2\int_{0}^{1}{1 - t^{\xi} \over 1 - t}\,\dd t - 2\int_{0}^{1}{1 - t^{\xi + 1/2} \over 1 - t}\,\dd t - \int_{0}^{1}{t^{\xi}\ln\pars{t} \over 1 - t}\,\dd t}\dd\xi \label{1}\tag{1} \end{align} Integral are evaluated as: $$ \left\{\begin{array}{rcl} \ds{\int_{0}^{1}{1 - t^{\xi} \over 1 - t}\,\dd t} & \ds{=} & \ds{\Psi\pars{\xi + 1} + \gamma} \\[1mm] \ds{\int_{0}^{1}{1 - t^{\xi + 1/2} \over 1 - t}\,\dd t} & \ds{=} & \ds{\Psi\pars{\xi + {3 \over 2}} + \gamma} \\[5mm] \ds{\int_{0}^{1}{t^{\xi}\ln\pars{t} \over 1 - t}\,\dd t} & \ds{=} & \ds{\left.-\,\partiald{}{\mu}\int_{0}^{1}{1 - t^{\mu} \over 1 - t}\,\dd t\, \right\vert_{\ \mu\ =\ \xi}} \\[1mm] & = & \ds{\left.-\,\partiald{\Psi\pars{\mu + 1}}{\mu} \,\right\vert_{\ \mu\ =\ \xi} = -\Psi\, '\pars{\xi + 1}} \end{array}\right. $$ (\ref{1}) becomes: \begin{align} &\bbox[5px,#ffd]{\sum_{n = 2}^{\infty} {\pars{-1}^{n}\,\zeta\pars{n} \over 2^{n - 1}\, n}} = \int_{0}^{\infty}\bracks{% 2\Psi\pars{\xi + 1} - 2\Psi\pars{\xi + {3 \over 2}} + \Psi\, '\pars{\xi + 1}}\dd\xi \\[5mm] = &\ \left.2\ln\pars{\Gamma\pars{\xi + 1} \over \Gamma\pars{\xi + 3/2}} + \Psi\pars{\xi + 1}\,\right\vert_{\ \xi\ =\ 0}^{\ \xi\ \to\ \infty} \\[5mm] = &\ \underbrace{\lim_{\xi \to \infty}\bracks{% 2\ln\pars{\Gamma\pars{\xi + 1} \over \Gamma\pars{\xi + 3/2}} + \Psi\pars{\xi + 1}}}_{\ds{=\ 0}}\ -\ \bracks{2\ln\pars{\Gamma\pars{1} \over \Gamma\pars{3/2}} + \Psi\pars{1}} \\[5mm] = &\ 2\ln\pars{{1 \over 2}\,\root{\pi}} + \gamma = \bbx{\gamma + \ln\pars{\pi \over 4}} \\ & \end{align} with $\ds{\Gamma\pars{1} = 1\,,\ \Gamma\pars{1/2} = \root{\pi}}$ and $\ds{\Psi\pars{1} = -\gamma}$.