How can I evaluate $\sum_{n=0}^\infty \frac{(2n)!}{(n!)^2(2n+1)4^n}$? 
Evaluate:
$$\sum_{n=0}^\infty \frac{(2n)!}{(n!)^2(2n+1)4^n}$$

I'm trying to solve this infinite series of problems, but I'm not making any progress. I'd appreciate your help.
I'm sorry if my writing is rude because I'm not good at English.
Thank you for reading.
 A: [I see you have brought a little change to your question].
Let $$f(x):=\sum_{n=0}^{\infty}\frac{x^n}{2n+1}\binom{2n}{n}.$$
(it will be time to change $x$ into $\frac14$ at the very last step).
Let:
$$g(x):=xf(x^2)=\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}\binom{2n}{n}$$
is a primitive function of
$$h(x):=\sum_{n=0}^{\infty}{(x^2)}^n\binom{2n}{n}=\dfrac{1}{\sqrt{1-4x^2}} \tag{1}$$
the last equality coming from the classical generating function for central binomial coefficients (Generating functions and central binomial coefficient).
It remains now to work backwards, the remaining difficulty being the integration of expression (1).
A: Your series can be written as $$S=\sum_{k=0}^{\infty} {2k\choose k} \frac{2^{-2k}}{2k+1}$$
Note that $$\sum_{k=0}^{\infty} {2k \choose k} x^{2k}=\frac{1}{\sqrt{1-4x^2}}$$
Integrate both sides w.r.t. $x$ from $x=0$ to $x=1/2$, then you get
$$\frac{1}{2}\sum_{k=0}^{\infty} {2k \choose k} \frac{2^{-2k}}{2k+1}=\int_{0}^{1/2} \frac{dx}{\sqrt{1-4x^2}}=\frac{1}{2}\sin^{-1} 2x|_{0}^{1/2}=\frac{\pi}{4}.$$
Finally,$$S=\pi/2$$
