calculus - 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}
$$

Answer

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 $(3)$.