Power series of $\pi$

Question

How do I prove that

$\displaystyle{\sum_{i=1}^{\infty}{\frac{2^n(n!)^2}{(2n)!}}=1+\frac{\pi}{2}}$

Answer 1

Challenging question. You may notice that $\frac{n!^2}{(2n)!}=\binom{2n}{n}^{-1}$, then prove that by integration by parts and induction we have $$ \int_{0}^{\pi/2}\sin(x)^{2n-1}\,dx = \frac{4^n}{2n\binom{2n}{n}} \tag{1}$$ It follows that

$$ S=\sum_{n\geq 1}2^n\binom{2n}{n}^{-1} = \int_{0}^{\pi/2}\sum_{n\geq 1}\frac{2n}{2^n}\sin(x)^{2n-1}\,dx = \int_{0}^{\pi/2}\frac{4\sin(x)}{(2-\sin^2 x)^2}\,dx.\tag{2}$$ Now it is not difficult to check that the last integral equals $\frac{\pi+2}{2}$: for instance, by integration by parts, since it equals: $$ S = \int_{0}^{1}\frac{4\,dt}{(1+t^2)^2}.\tag{3}$$

Answer 2

We have that $$\sum_{n=1}^{\infty}\frac{2^n(n!)^2}{(2n)!}=\sum_{n=1}^{\infty}\frac{4^n (1/2)^n}{\binom{2n}{n}}=\sum_{n=0}^{\infty}\frac{4^n (1/2)^n}{\binom{2n}{n}}-1\\=Z(1/2)-1=\left(2\cdot\frac{\pi}{4}+2\right)-1=\frac{\pi}{2}+1$$ where we used the fact that for $|t|<1$, $$Z(t)=\sum_{n=0}^{\infty}\frac{4^n t^n}{\binom{2n}{n}}= \frac{1}{1-t}\sqrt{\frac{t}{1-t}}\arctan \sqrt{\frac{t}{1-t}}+\frac{1}{1-t}.$$ A proof is given by Theorem 2.1 here.

Answer 3

One way to deal with this is notice the factor in the summand is proportional to a Beta function.

$$\frac{n!^2}{(2n)!} = \frac{n}{2}\frac{\Gamma(n)\Gamma(n)}{\Gamma(2n)} = \frac{n}{2}\int_0^1 t^{n-1}(1-t)^{n-1} dt$$

This leads to $$\sum_{n=1}^\infty \frac{2^n n!^2}{(2n)!} = \int_0^1 \left(\sum_{n=1}^\infty n (2t(1-t))^{n-1}\right) dt = \int_0^1 \frac{dt}{(1-2t(1-t))^2} = 4\int_0^1 \frac{dt}{(1+(2t-1)^2)^2} $$ Change variable to $2t-1 = \tan\theta$, the series reduces to

$$2\int_{-\pi/4}^{\pi/4} \frac{d\theta}{1+\tan^2\theta} = \int_{-\pi/4}^{\pi/4} (1 + \cos2\theta) d\theta = \left[\theta + \frac12\sin(2\theta)\right]_{-\pi/4}^{\pi/4} = \frac{\pi}{2} + 1 $$