Evaluating series of zeta values like $\sum_{k=1}^{\infty} \frac{\zeta(2k)}{k16^{k}}=\ln(\pi)-\frac{3}{2}\ln(2) $

Question

Somehow I derived these values a few years ago but I forgot how. It cannot be very hard (certainly doesn't require "advanced" knowledge) but I just don't know where to start.

Here are the sums: $$ \begin{align} \sum_{k=1}^{\infty} \frac{\zeta(2k)}{4^{k}}&=\frac{1}{2} \\\\ \sum_{k=1}^{\infty} \frac{\zeta(2k)}{16^{k}}&=\frac{4-\pi}{8} \\\\ \sum_{k=1}^{\infty} \frac{\zeta(2k)}{k4^{k}}&=\ln(\pi)-\ln(2) \\\\ \sum_{k=1}^{\infty} \frac{\zeta(2k)}{k16^{k}}&=\ln(\pi)-\frac{3}{2}\ln(2). \end{align} $$

Answer 1

Hint. One may start with the classic series expansion, which may come from the Weierstrass infinite product of the sine function,

$$ \sum _{n=1}^{\infty } \frac{x^2}{n^2+x^2}=\frac{1}{2} (-1+\pi x \cot (\pi x)) , \quad|x|<1. \tag1 $$

Expanding the left hand side of $(1)$ one deduces

$$ \sum_{k=1}^{\infty } \zeta(2k)\:x^{2k}=\frac{1}{2} (1-\pi x \cot (\pi x)) , \quad|x|<1. \tag2 $$

By dividing $(2)$ by $x$ and integrating one gets

$$ \sum_{k=1}^{\infty } \zeta(2k)\:\frac{x^{2k}}k=\log \left(\frac{\pi x}{\sin(\pi x)}\right) , \quad|x|<1. \tag3 $$

Your equalities are now obtained by putting $x:=\dfrac12,\, \dfrac14$ in $(2)$ and in $(3)$.

Answer 2

Hint: $\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} = \sum_{n=1}^{\infty} \sum_{k=1}^{\infty}$

This is justified by Tonelli's theorem, for instance. In the first one, the inner sum becomes a geometric series, and when you compute it, the whole sum becomes a telescoping series:

$$\frac12 \sum \left( \frac1{2n -1} - \frac1{2n + 1} \right)$$

Etc.

Answer 3

The first thing that comes to mind is changing the order of summation. For the first one,

$$ \eqalign{\sum_{k=1}^\infty \dfrac{\zeta(2k)}{4^k} &= \sum_{n=1}^\infty \sum_{k=1}^\infty \dfrac{1}{(4n^2)^k }\cr &= \sum_{n=1}^\infty \dfrac{1}{4n^2-1}\cr &= \sum_{n=1}^\infty \left(\frac{1/2}{2n-1} - \frac{1/2}{2n+1}\right) = \frac{1}{2}}$$

EDIT: Somewhat more generally, for $r > 1$

$$ \sum_{k=1}^\infty \dfrac{\zeta(2k)}{r^k} = \sum_{n=1}^\infty \dfrac{1}{rn^2-1} = \dfrac{1}{2} - \dfrac{\pi\cot(\pi/\sqrt{r})}{2 \sqrt{r}}$$

$$ \eqalign{\sum_{k=1}^\infty \dfrac{\zeta(2k)}{k r^k} &= \int_r^\infty ds\; \sum_{k=1}^\infty \dfrac{\zeta(2k)}{s^{k+1}}\cr &= \int_r^\infty \dfrac{ds}{s} \left( \dfrac{1}{2} - \dfrac{\pi\cot(\pi/\sqrt{s})}{2 \sqrt{s}}\right) \cr &= \ln\left(\pi/\sqrt{r}\right) - \ln \left(\sin(\pi/\sqrt{r})\right) }$$

Answer 4

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Hereafter, $\ds{\Psi}$ and $\ds{\gamma}$ are the Digamma Function and the Euler-Mascheroni Constant, respectively.

With the well known identity $\ds{\left.\vphantom{\Large A}\Psi\pars{1 + z}\,\right\vert_{\ \verts{z}\ <\ 1} = -\gamma + \sum_{n = 2}^{\infty}\pars{-1}^{n}\,\zeta\pars{n}z^{n - 1}}$, we can show that \begin{align} &\fbox{$\ds{\ \sum_{n = 1}^{\infty}\zeta\pars{2n}z^{2n}\ }$} = \half\,z\bracks{\Psi\pars{1 + z} - \Psi\pars{1 - z}} = \half\,z\bracks{\Psi\pars{z} + {1 \over z} - \Psi\pars{1 - z}} \\[3mm] = &\ \fbox{$\ds{\ \half - {\pi \over 2}\,z\cot\pars{\pi z}\ }$} \quad\imp\quad \left\lbrace\begin{array}{lcl} \ds{\color{#f00}{\sum_{n = 1}^{\infty}{\zeta\pars{2n} \over 4^{n}}} = \color{#f00}{\half}} & \mbox{with} & \ds{z = \half} \\[2mm] \ds{\color{#f00}{\sum_{n = 1}^{\infty}{\zeta\pars{2n} \over 16^{n}}} = \color{#f00}{{4 - \pi \over 8}}} & \mbox{with} & \ds{z = {1 \over 4}} \end{array}\right. \end{align}

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Moreover, \begin{align} \int_{0}^{z}2\sum_{n = 1}^{\infty}\zeta\pars{2n}x^{2n - 1}\,\dd x & = \int_{0}^{z}\bracks{{1 \over x} - \pi\cot\pars{\pi x}}\,\dd x \\[3mm] \imp \sum_{n = 1}^{\infty}{\zeta\pars{2n} \over n}\,z^{2n} & = \ln\pars{\pi z} - \ln\pars{\sin\pars{\pi z}} \\[3mm] \imp & \left\lbrace\begin{array}{lcl} \ds{\color{#f00}{\sum_{n = 1}^{\infty}{\zeta\pars{2n} \over n\, 4^{n}}} = \color{#f00}{\ln\pars{\pi} - \ln\pars{2}}} & \mbox{with} & \ds{z = \half} \\[2mm] \ds{\color{#f00}{\sum_{n = 1}^{\infty}{\zeta\pars{2n} \over n\, 16^{n}}} = \color{#f00}{\ln\pars{\pi} - {3 \over 2}\,\ln\pars{2}}} & \mbox{with} & \ds{z = {1 \over 4}} \end{array}\right. \end{align}