How to prove $\sum_{n=0}^\infty \frac{(2n)!}{(2^nn!(2n+1))^2} = \frac{\pi}{2}\ln(2)$?

Question

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.

Answer 1

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

$$I(1/4)=-\int_0^{\pi/2}\ln \cos t dt=\frac{\pi\ln 2}{2}$$

Answer 2

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: $$\begin{align} \int_0^1 \frac{\arcsin(t)}tdt&=\left[\arcsin(t)\log(t)\right]_0^1- \int_0^1 \frac{\log(t)}{\sqrt{1-t^2}}dt\\ &=-\int_0^\pi \log(\sin u)du=\pi\log(2). \end{align}$$

Answer 3

$$\frac{(2n)!}{(2^nn!(2n+1))^2}=\frac{(2n)!}{2^{2n}(n!)^2(2n+1)^2}={2n\choose n}\frac{1}{2^{2n}(2n+1)^2}={2n\choose n}\frac{1}{(2n+1)^2}\left(\frac12\right)^{2n}$$ Now try and find a function whose taylor series fits this summation