How to prove $\sum_{n=0}^\infty \frac{(2n)!}{(2^nn!(2n+1))^2} = \frac{\pi}{2}\ln(2)$? I was messing around with some integrals and series when I arrived at the result (and WolframAlpha agrees):
$$\sum_{n=0}^\infty \frac{(2n)!}{(2^nn!(2n+1))^2} = \sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{1}{(2n+1)^2} = \frac{\pi}{2}\ln(2)$$
I therefore have a proof, but I would like to see how a proof would be done the other way around (i.e. starting with this series). Notice that the series could also be expressed with double factorials:
$$\sum_{n=0}^\infty \frac{(2n)!}{(2^nn!(2n+1))^2} = \sum_{n=0}^\infty \frac{(2n-1)!!}{(2n)!!}\frac{1}{(2n+1)^2}$$
My Derivation
My derivation is quite simple. I was considering (and trying to solve) the integral
$$\int_0^1 \frac{\arcsin(x)}{x} \, dx$$
I couldn't work it out, however I found the exact same question here on math exchange and the answers show that the integral equals $\frac{\pi}{2}\ln(2)$. What I did found however, was that I could plug the Taylor series for $\arcsin(x)$ (the proof of the taylor series can be found here) into the original series to only be left with a series. Here's the Taylor series for $\arcsin(x)$:
$$
  \arcsin(x) = \sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{ x^{2n+1}}{2n+1}
$$
So plugging in the series above into the integral gives us:
$$\int_0^1 \frac{1}{x} \sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{ x^{2n+1}}{2n+1} \, dx$$
Now here we have to be careful, but since the series works on the integral's interval $[0,1]$ and the summand is only positive, we can interchange the the integral and the sum, leaving us with:
$$\sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{1}{2n+1} \, \int_0^1 \frac{x^{2n+1}}{x} \, dx$$
And then solving the integral in the sum gives us the final result:
$$\sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{1}{(2n+1)^2} = \frac{\pi}{2}\ln(2)$$
And then expanding the binomial gives the original series in the title.
 A: Consider the well-known series $$f(x)=\sum_{n=0}^{\infty}{2n\choose n}x^n=\frac{1}{\sqrt{1-4x}}$$
We will attempt to evaluate the series
$$I(x)=\sum_{n=0}^{\infty}{2n\choose n}\frac{x^n}{(2n+1)^2}$$
A simple relation between these two series exists:
$$\frac{d^2}{dt^2}[e^tI(e^{2t})]=e^t f(e^{2t})=\frac{e^t}{\sqrt{1-4e^{2t}}}$$
To obtain our goal series we just need to integrate twice with an appropriate boundary condition
$$e^t I(e^{2t})=\int_{-\infty}^tdx\int_{-\infty}^{x}dT\frac{e^T}{\sqrt{1-4e^{2T}}}$$
The inner integral is easy to do which leads us to the expression
$$I(x)=\frac{1}{2\sqrt{x}}\int_{-\infty}^{\frac{\ln x}{2}}dt ~ \arcsin(2e^t)$$
which can be further simplified to
$$I(x)=\frac{1}{2\sqrt{x}}\int_{0}^{2\sqrt{x}}\frac{\arcsin u}{u}du$$
Our goal is to evaluate this at $x=1/4$. Performing an integration by parts and afterwards the substitution $u=\cos t$ we reduce the entire thing to a well-known integral that can be calculated in various ways
$$I(1/4)=-\int_0^{\pi/2}\ln \cos t dt=\frac{\pi\ln 2}{2}$$
A: Start with well-known series:
$$
f(x)=\sum_{k=0}^\infty\binom{2n}n\left(\frac x2\right)^{2n}=\frac1{\sqrt{1-x^2}}\tag1
$$
From this one obtains:
$$\begin{align}
&\sum_{k=0}^\infty\binom{2n}n\frac1{2n+1}\left(\frac x2\right)^{2n}=\frac1x\int_0^x \frac{dt}{\sqrt{1-t^2}}=\frac{\arcsin(x)}x\\
&\sum_{k=0}^\infty\binom{2n}n\frac1{(2n+1)^2}\left(\frac 12\right)^{2n}=\frac12\int_0^1 \frac{\arcsin(t)}tdt.\\
\end{align}$$
The last integral can be evaluated as:
$$\begin{align}
\int_0^1 \frac{\arcsin(t)}tdt&=\left[\arcsin(t)\log(t)\right]_0^1-
\int_0^1 \frac{\log(t)}{\sqrt{1-t^2}}dt\\
&=-\int_0^\pi \log(\sin u)du=\pi\log(2).
\end{align}$$
A: $$\frac{(2n)!}{(2^nn!(2n+1))^2}=\frac{(2n)!}{2^{2n}(n!)^2(2n+1)^2}={2n\choose n}\frac{1}{2^{2n}(2n+1)^2}={2n\choose n}\frac{1}{(2n+1)^2}\left(\frac12\right)^{2n}$$
Now try and find a function whose taylor series fits this summation
