Evaluating series of zeta values like $\sum_{k=1}^{\infty} \frac{\zeta(2k)}{k16^{k}}=\ln(\pi)-\frac{3}{2}\ln(2) $ 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:
$$
\begin{align}
\sum_{k=1}^{\infty} \frac{\zeta(2k)}{4^{k}}&=\frac{1}{2}
\\\\
\sum_{k=1}^{\infty} \frac{\zeta(2k)}{16^{k}}&=\frac{4-\pi}{8}
\\\\
\sum_{k=1}^{\infty} \frac{\zeta(2k)}{k4^{k}}&=\ln(\pi)-\ln(2)
\\\\
\sum_{k=1}^{\infty} \frac{\zeta(2k)}{k16^{k}}&=\ln(\pi)-\frac{3}{2}\ln(2).
\end{align}
$$
 A: 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 $(3)$.
A: 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.
A: 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 $r > 1$
$$ \sum_{k=1}^\infty \dfrac{\zeta(2k)}{r^k} = \sum_{n=1}^\infty \dfrac{1}{rn^2-1} = \dfrac{1}{2} - \dfrac{\pi\cot(\pi/\sqrt{r})}{2 \sqrt{r}}$$
$$ \eqalign{\sum_{k=1}^\infty \dfrac{\zeta(2k)}{k r^k} &= \int_r^\infty ds\; \sum_{k=1}^\infty \dfrac{\zeta(2k)}{s^{k+1}}\cr
&= \int_r^\infty \dfrac{ds}{s} \left( \dfrac{1}{2} - \dfrac{\pi\cot(\pi/\sqrt{s})}{2 \sqrt{s}}\right) \cr
&=   \ln\left(\pi/\sqrt{r}\right) - \ln \left(\sin(\pi/\sqrt{r})\right) }$$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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}

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}
