Why does this series $\sum_{n=0}^{\infty} \frac{(n!)^{2}}{(2n)!}$ converge?

Question

The following series $$\sum_{n=0}^{\infty} \frac{(n!)^{2}}{(2n)!}$$ converges. It fails the divergence test, but once I apply the ratio test, the limit is always equal to $\infty$. Unless you cannot distribute an exponent on a factorial product, I don't see what I am doing wrong.

I tried using Stirling's approximation to construct a geometric series, but I always got $1$ for the ratio, so the test remains inconclusive.

As for the other tests, I didn't find any similar series against which to compare this one to (the limit comparison always gave me $\infty$ or $0$...).

Any help or tips on how to take a different spin on the problem are greatly appreciated.

Answer 1

We have the following series:

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

To prove convergence, use the Ratio Test:

$$\lim_{n\to\infty}\left|\frac{\left((n+1)!\right)^{2}}{(2(n+1))!}*\frac{(2n)!}{(n!)^{2}}\right|$$ $$\lim_{n\to\infty}\left|\frac{(n+1)!(n+1)!}{(2n+2)!}*\frac{(2n)!}{(n!)(n!)}\right|$$ $$\lim_{n\to\infty}\left|\frac{(n+1)(n+1)}{(2n+1)(2n+2)}\right|$$

As you notice, the leading degree of the denominator is the same as the leading degree of the numerator. Therefore, we look at the coefficients, and so the limit goes to $\frac{1}{4}$, and the series convergences absolutely by the Ratio Test.

Answer 2

By Stirling:

$$\frac{(n!)^2}{(2 n)!} \sim \frac{n^{2 n} e^{-2 n} 2 \pi n}{(2 n)^{2 n} e^{-2 n} \sqrt{2 \pi 2 n}} = \frac{\sqrt{2 \pi n}}{\sqrt{2} 2^{2 n}}$$

so the sum certainly converges by comparison with

$$\sum_{n=0}^{\infty} \sqrt{n} \, 2^{-2 n} $$

Answer 3

You can conclude by comparison, since $$\frac{(n!)^2}{(2n)!} \sim_{n\to\infty} \frac{\sqrt{\pi n}}{2^{2n}}$$ by Stirling's approximation.

Answer 4

If you apply the ratio test you should get the limit of a(n+1)/a(n) tends to 1/4 hence we get convergence.

Answer 5

The ratio test gives you $$\frac{(2n)!}{(n!)^2} \cdot \frac{((n+1)!)^2}{(2n+2)!} = \frac{(n+1)(n+1)}{(2n+2)(2n+1)} \to \frac 14 < 1$$ as $n$ tends to infinity. So the series is convergent.

Answer 6

You can use the root test.

$$\sqrt[n]{\frac{(n!)^2}{(2n)!}}=\sqrt[n]{\frac{2\cdot 3\cdots n}{(n+1)\cdots(2n)}}\leq \frac{1}{n}\sum_{j=1}^n \frac{j}{n+j}\leq \frac{1}{n} \sum_{j=1}^{n}\frac{j}{j+j}=\frac{1}{2},$$(by AM-GM inequality) so$$\limsup_{n\to \infty} \sqrt[n]{\frac{(n!)^2}{(2n)!}} \leq \frac{1}{2}.$$ Therefore the series converges.

Answer 7

$$\begin{eqnarray*}\sum_{n\geq 0}\frac{n!^2}{(2n)!}=\sum_{n\geq 0}\frac{\Gamma(n+1)^2}{\Gamma(2n+1)}&=&\sum_{n\geq 0}(2n+1)\int_{0}^{1}x^n(1-x)^n\,dx\\&=&\int_{0}^{1}\frac{1+x-x^2}{(1-x+x^2)^2}\,dx\\&=&\left.\frac{2}{3}\cdot\frac{2x-1}{1-x+x^2}+\frac{2}{3\sqrt{3}}\cdot\arctan\left(\frac{2x-1}{\sqrt{3}}\right)\right|_{0}^{1}\\&=&\color{red}{\frac{2}{27}(18+\pi\sqrt{3}).}\end{eqnarray*}$$

Answer 8

For $n≥1$, easy to know: $\frac{n!^2}{(2n)!}\le\frac{1}{2^n}$, hence the answer is obvious.