TITLE: Evaluating a sum involving binomial coefficient in denominator
QUESTION [21 upvotes]: I came across the following sum:
$$\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^2}\frac{4^k}{{2k \choose k}}$$

I thought that this can be evaluated using the expansion of $\dfrac{\sin^{-1}x}{\sqrt{1-x^2}}$ but I couldn't make any use of it. Then I tried to use the following:
$$\frac{1}{(2k+1)}\frac{1}{{2k \choose k}}=\frac{\Gamma(k+1)\Gamma(k+1)}{\Gamma(2k+2)}=\int_0^1 x^k(1-x)^k\,dx$$
but this didn't help either and now I am stuck. 
Any help is greatly appreciated. Thanks!

EDIT:
The sum originated from the following definite integral:
$$\int_0^{\pi/2}\tan^{-1}(\sin x)\,dx$$

REPLY [12 votes]: Determine a Related Generating Function
Using the Beta function, we get the following identity:
$$
\frac1{\binom{2n}{n}}=(2n+1)\int_0^1t^n(1-t)^n\mathrm{d}t\tag{1}
$$
Thus,
$$
\begin{align}
\sum_{n=0}^\infty\frac{(-4)^nx^{2n}}{(2n+1)\binom{2n}{n}}
&=\int_0^1\frac1{1+4x^2t(1-t)}\mathrm{d}t\tag{2a}\\
&=\int_0^1\frac1{1+x^2-x^2(2t-1)^2}\mathrm{d}t\tag{2b}\\
&=\frac1{1+x^2}\int_0^1\frac1{1-\frac{x^2}{1+x^2}(2t-1)^2}\mathrm{d}t\tag{2c}\\
&=\frac1{1+x^2}\int_{-1}^1\frac1{1-\frac{x^2}{1+x^2}t^2}\frac12\mathrm{d}t\tag{2d}\\
&=\frac1{2x\sqrt{1+x^2}}\int_{-x/\sqrt{1+x^2}}^{x/\sqrt{1+x^2}}\frac1{1-t^2}\mathrm{d}t\tag{2e}\\
&=\frac1{x\sqrt{1+x^2}}\mathrm{arctanh}\left(\frac{x}{\sqrt{1+x^2}}\right)\tag{2f}\\
&=\frac1{x\sqrt{1+x^2}}\mathrm{arcsinh}(x)\tag{2g}
\end{align}
$$
Explanation:
$\text{(2a)}$: multiply $(1)$ by $\frac{(-4)^nx^{2n}}{2n+1}$ and sum using the formula for the sum of a geometric series
$\text{(2b)}$: rearrange the denominator of the integrand
$\text{(2c)}$: divide numerator and denominator by $1+x^2$
$\text{(2d)}$: substitute $t\mapsto(t+1)/2$
$\text{(2e)}$: substitute $t\mapsto t\sqrt{1+x^2}\,/x$
$\text{(2f)}$: integrate
$\text{(2g)}$: $\tanh(x)=\left.\sinh(x)\middle/\sqrt{1+\sinh^2(x)}\right.$

Integrate the Generating Function to Evaluate the Sum
$$
\begin{align}
\sum_{n=0}^\infty\frac{(-4)^n}{(2n+1)^2\binom{2n}{n}}
&=\int_0^1\frac1{x\sqrt{1+x^2}}\mathrm{arcsinh}(x)\,\mathrm{d}x\tag{3a}\\
&=-\int_0^1\mathrm{arcsinh}(x)\frac1{\sqrt{\vphantom{\big|}1+1/x^2}}\mathrm{d}\frac1x\tag{3b}\\
&=-\int_0^1\mathrm{arcsinh}(x)\,\mathrm{d}\,\mathrm{arcsinh}\left(\frac1x\right)\tag{3c}\\
&=-\,\mathrm{arcsinh}^2(1)+\int_0^1\mathrm{arcsinh}\left(\frac1x\right)\,\mathrm{d}\,\mathrm{arcsinh}(x)\tag{3d}\\
&=-\,\mathrm{arcsinh}^2(1)-\int_1^\infty\mathrm{arcsinh}(x)\,\mathrm{d}\,\mathrm{arcsinh}\left(\frac1x\right)\tag{3e}\\
&=-\,\mathrm{arcsinh}^2(1)+\int_1^\infty\frac1{x\sqrt{1+x^2}}\mathrm{arcsinh}(x)\,\mathrm{d}x\tag{3f}\\
&=-\frac12\,\mathrm{arcsinh}^2(1)+\frac12\int_0^\infty\frac1{x\sqrt{1+x^2}}\mathrm{arcsinh}(x)\,\mathrm{d}x\tag{3g}\\
&=-\frac12\,\mathrm{arcsinh}^2(1)+\frac12\int_0^\infty\frac{t\,\mathrm{d}t}{\sinh(t)}\tag{3h}
\end{align}
$$
Explanation:
$\text{(3a)}$: integrate $\text{(2g)}$
$\text{(3b)}$: pull a factor of $x$ from the square root
$\text{(3c)}$: prepare to integrate by parts
$\text{(3d)}$: integrate by parts
$\text{(3e)}$: substitute $x\mapsto1/x$
$\text{(3f)}$: $\mathrm{d}\,\mathrm{arcsinh}\left(\frac1x\right)=-\frac1{x\sqrt{1+x^2}}\,\mathrm{d}x$
$\text{(3g)}$: average $\text{(3a)}$ and $\text{(3f)}$
$\text{(3h)}$: substitute $x=\sinh(t)$
Expand $\frac1{\sinh(x)}$ as a series in $e^{-kt}$:
$$
\begin{align}
\int_0^\infty\frac{t\,\mathrm{dt}}{\sinh(t)}
&=\int_0^\infty\sum_{k=0}^\infty2t\,e^{-(2k+1)t}\,\mathrm{d}t\\
&=\sum_{k=0}^\infty\frac2{(2k+1)^2}\\
&=\frac{\pi^2}4\tag{4}
\end{align}
$$
Combining $(3)$ and $(4)$ yields
$$
\begin{align}
\sum_{n=0}^\infty\frac{(-4)^n}{(2n+1)^2\binom{2n}{n}}
&=\frac{\pi^2}8-\frac12\mathrm{arcsinh}^2(1)\tag{5}
\end{align}
$$