Finding $\sum_{n=0}^∞ \frac{(n!)^2}{(2n)!}$

Question

Find the value of the follwing series: $$\sum_{n=0}^\infty a_n :=\sum_{n=0}^∞ \frac{(n!)^2}{(2n)!}.$$

The series clearly converges, as can be seen with the quotient test -

$$ \lim_{n\to\infty} \frac{(n+1)!^2/(2n+2)!}{n!^2/(2n)!}=\lim_{n\to\infty}\frac{n+1}{4n+2}=\frac 1 4.$$

As for the value, Wolfram Alpha suggests that it is $$\sum_{n=0}^\infty a_n=\frac 4 3 + \frac {2\pi}{9\sqrt 3},$$ which is calculated using hypergeometric functions. I am trying to understand the way of calculation (which can hopefully be simplified a bit). We note that $a_n = {2n\choose n}^{-1}$, and that, as proven here, we have an "almost reciprocal" Taylor series $\arcsin(x) = \sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{ x^{2n+1}}{2n+1}$. There is no general "strong" connection between $\sum a_n x^n$ and $\sum \frac{1}{a_n} x^n$, as mentioned here; but the case of binomial coefficients is perhaps easier to work with. Furthermore, a similar calculation yields

$$ \sum_{n=0}^\infty \frac{(n!)^2}{(2n)!} \left( \frac 1 2 \right)^n = \frac 8 {49} \left( 7+\sqrt 7 \arcsin\frac 1 {2\sqrt 2} \right).$$

So, I hope there is a relatively simple method - perhaps only using simple properties of hypergeometric function - for calculating the series $\sum_{n=0}^\infty \frac{(n!)^2}{(2n)!} x^n $.

Answer 1

Using the beta distribution integral $$\int_{0}^{1}x^a(1-x)^b\,dx = \dfrac{a! \cdot b!}{(a+b+1)!},$$ we have $$\int_{0}^{1}(2n+1)x^n(1-x)^n\,dx = (2n+1)\dfrac{n! \cdot n!}{(2n+1)!} = \dfrac{(n!)^2}{(2n)!}.$$

Hence, we have $$\begin{align*}\sum_{n = 0}^{\infty}\dfrac{(n!)^2}{(2n)!} &= \sum_{n = 0}^{\infty}\int_{0}^{1}(2n+1)x^n(1-x)^n\,dx \\ &= \int_{0}^{1}\sum_{n = 0}^{\infty}(2n+1)(x(1-x))^n \\ &= \int_{0}^{1}\dfrac{1+x(1-x)}{(1-x(1-x))^2}\,dx \\ &= \int_{0}^{1}\dfrac{1+x-x^2}{(1-x+x^2)^2}\,dx\end{align*},$$ where the interchange of the summation and integral is justified since the integrand is positive everywhere, and we have used the series $\displaystyle\sum_{n = 0}^{\infty}(2n+1)y^n = \dfrac{1+y}{(1-y)^2}$ for $|y| < 1$.

Finally, you can use partial fraction decomposition to get $$\int\dfrac{1+x-x^2}{(1-x+x^2)^2}\,dx = \dfrac{\tfrac{2}{3}(2x-1)}{x^2-x+1}+\dfrac{2}{3\sqrt{3}}\arctan\left(\dfrac{2x-1}{\sqrt{3}}\right)+C,$$ and then evaluate to get $$\int_{0}^{1}\dfrac{1+x-x^2}{(1-x+x^2)^2}\,dx = \dfrac{4}{3} + \dfrac{2\pi}{9\sqrt{3}}.$$

Similarly, you can obtain $$\sum_{n = 0}^{\infty}\dfrac{(n!)^2}{(2n)!}a^n = \int_{0}^{1}\dfrac{1+ax-ax^2}{(1-ax+ax^2)^2}\,dx,$$ for $|a| < 1$ and use similar techniques to evaluate it.