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.
 A: Note that $$\frac{1+(-1)^k}{2} = \begin{cases}1 &\text{if $k$ is even}\\ 0 &\text{if $k$ is odd}\end{cases}$$
implies that $$\sum_k a_{2k} = \sum_k \frac{1+(-1)^k}{2} a_k.$$  Taking $a_k = \frac{\binom{2k}{k}}{(k/2)^2 16^{k/2}}$ yields
\begin{align}
\sum_{k=1}^\infty \frac{\binom{4k}{2k}}{k^2 16^k}
&= \sum_{k=1}^\infty \frac{1+(-1)^k}{2}\frac{\binom{2k}{k}}{(k/2)^2 16^{k/2}} \\
&= 2\sum_{k=1}^\infty (1+(-1)^k)\frac{\binom{2k}{k}}{k^2 4^k} \\
&= 2\sum_{k=1}^\infty \frac{\binom{2k}{k}}{k^2 4^k} + 2\sum_{k=1}^\infty (-1)^k\frac{\binom{2k}{k}}{k^2 4^k} \\
\end{align}
A: Using @RobPratt's elegant solution
$$S_1(x)=\sum_{k=1}^\infty \frac{\binom{2k}{k}}{k^2 }x^k=2 x \, _4F_3\left(1,1,1,\frac{3}{2};2,2,2;4 x\right)$$ Simplifying
$$S_1(x)=2 x \left(\frac{\text{Li}_2\left(\frac{1}{2}-\frac{1}{2} \sqrt{1-4
   x}\right)}{x}-\frac{\left(\log \left(1+\sqrt{1-4 x}\right)-\log (2)\right)^2}{2
   x}\right)$$
$$S_1(4)=2 \text{Li}_2\left(\frac{1}{2} \left(1-i \sqrt{15}\right)\right)+\left(\tan
   ^{-1}\left(\sqrt{15}\right)-i \log (2)\right)^2$$
$$\color{blue}{S_1(4)=\frac {\pi^2}6-2 \log ^2(2)}$$
$$S_2(x)=\sum_{k=1}^\infty (-1)^k \frac{\binom{2k}{k}}{k^2 }x^k=-2 x \, _4F_3\left(1,1,1,\frac{3}{2};2,2,2;-4 x\right)$$ Simplifying
$$S_2(x)=-2 x \left(\frac{\left(\log \left(1+\sqrt{1+4 x}\right)-\log (2)\right)^2}{2
   x}-\frac{\text{Li}_2\left(\frac{1}{2}-\frac{1}{2} \sqrt{1+4 x}\right)}{x}\right)$$
$$S_2(4)=2 \text{Li}_2\left(\frac{1}{2} \left(1-\sqrt{17}\right)\right)-\log ^2(2)-\log
   \left(1+\sqrt{17}\right) \text{csch}^{-1}(4)$$
$$\color{blue}{S_2(4)=2 \text{Li}_2\left(\frac{1}{2}-\frac{1}{\sqrt{2}}\right)-\left(\sinh ^{-1}(1)-\log (2)\right)^2}$$
Combining all of the above
$$\color{red}{2(S_1(4)+S_2(4))=4 \text{Li}_2\left(\frac{1}{2}-\frac{1}{\sqrt{2}}\right)+\frac{\pi ^2}{3}-6 \log
   ^2(2)+2 \sinh ^{-1}(1) \left(2 \log (2)-\sinh ^{-1}(1)\right)}$$
A: 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
$$I_+ = \int_0^1 \frac{f_+(x)}{x} \, {\rm d}x \stackrel{u=\sqrt{1+\sqrt{x}}}{=} 8 \int_1^\sqrt{2} \frac{\log(2)-\log(1+u)}{u^2-1} \, u \, {\rm d}u \\
\stackrel{t=\frac{u-1}{u+1}}{=} \int_0^{\frac{\sqrt{2}-1}{\sqrt{2}+1}} \left( \frac{4\log(1-t)}{t} + \frac{8\log(1-t)}{1-t} \right) {\rm d}t = -4\,{\rm Li}_2\left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right) - 4 \log^2\left(\frac{2}{\sqrt{2}+1}\right) \, .$$
