Evaluating $\sum \limits_{k=1}^{\infty} \frac{\binom{4k}{2k}}{k^2 16^k}$ I want to find the closed form of:
$\displaystyle \tag*{}\sum \limits_{k=1}^{\infty} \frac{\binom{4k}{2k}}{k^2 16^k}$
I tried to use the taylor expansion of $\frac{1}{\sqrt{1-x}}$ and $\frac{1}{\sqrt{1-4x}}$ but both of them had $\binom{2n}{n}$ in the numerator and no square in the denominator. I am unsuccesful even in using $\arcsin^2x$ expansion.
However, we know a well known result:
$\displaystyle \tag*{} \sum \limits_{k=1}^{\infty} \frac{\binom{2n}{n}}{k^24^k} = \frac{\pi^2}{6} - 2\ln^2(2)$
So this somehows tells (?) maybe we can decompose our sum into two parts with one being Basel sum. Any help would be appreciated, thanks.

If you accept that $$f(x)=\sum_{k=1}^\infty \frac{x^{k}}{k\,16^k} \binom{4k}{2k} = f_+(x) + f_-(x)$$
where
$$f_{\pm}(x) = 2\log(2) - 2 \log\left(1+\sqrt{1\pm\sqrt{x}}\right) \, ,$$
which shouldn't be too difficult to obtain (see RobPratt), then
$$I_- =\int_0^1 \frac{f_-(x)}{x} \, {\rm d}x \stackrel{u=\sqrt{1-\sqrt{x}}}{=} 8 \int_0^1 \frac{\log(2)-\log(1+u)}{1-u^2} \, u \, {\rm d}u \\
\stackrel{t=\frac{1-u}{1+u}}{=} \int_0^1 \left( \frac{4\log(1+t)}{t} - \frac{8\log(1+t)}{1+t} \right) {\rm d}t = \frac{\pi^2}{3} - 4\log^2(2)$$
and
$$I_+ = \int_0^1 \frac{f_+(x)}{x} \, {\rm d}x \stackrel{u=\sqrt{1+\sqrt{x}}}{=} 8 \int_1^\sqrt{2} \frac{\log(2)-\log(1+u)}{u^2-1} \, u \, {\rm d}u \\
\stackrel{t=\frac{u-1}{u+1}}{=} \int_0^{\frac{\sqrt{2}-1}{\sqrt{2}+1}} \left( \frac{4\log(1-t)}{t} + \frac{8\log(1-t)}{1-t} \right) {\rm d}t = -4\,{\rm Li}_2\left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right) - 4 \log^2\left(\frac{2}{\sqrt{2}+1}\right) \, .$$
