Determine if this series $\sum\limits_{n=1}^\infty \frac{(n!)^2}{(2n)!}$ converges or diverges, and prove your answer?

Question

Determine if this series $$\sum\limits_{n=1}^\infty \frac{(n!)^2}{(2n)!}$$ converges or diverges, and prove your answer?

I've been able to prove similar problems, but I'm confused now that there's a factorial involved. Can someone help me out here?

Answer 1

Let $a_n$ be the $n$=th term. We use the Ratio Test, and calculate $\lim_{n\to\infty} \frac{a_{n+1}}{a_n}$. We have $$\frac{a_{n+1}}{a_n}=\frac{ \frac{((n+1)!)^2}{(2n+2)!}}{\frac{(n!)^2}{(2n)!}}=\frac{((n+1)!)^2 (2n)!}{(n!)^2(2n+2)!}$$ Now we start to simplify. Note that $\frac{(n+1)!}{n!}=n+1$ and $\frac{(2n)!}{(2n+2)!}=\frac{1}{(2n+1)(2n+2)}$, so our ratio simplifies to $$\frac{(n+1)^2}{(2n+1)(2n+2)},$$ which further simplifies to $$\frac{n+1}{2(2n+1)}.$$ Now find the limit as $n\to\infty$, perhaps by dividing top and bottom by $n$. The limit is $\frac{1}{4}\lt 1$, so we have convergence.

Answer 2

We apply the ratio test.

$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}= \lim_{n\to\infty}\dfrac{\frac{((n+1)!)^2}{(2(n+1))!}}{\frac{(n!)^2}{(2n)!}}= \lim_{n\to\infty}\dfrac{((n+1)!)^2(2n)!}{(2n+2)!\cdot (n!)^2}\\=\lim_{n\to\infty}\dfrac{((n+1)(n!))^2\cdot (2n)!}{(2n+2)(2n+1)(2n)!\cdot (n!)^2}=\lim_{n\to\infty}\dfrac{((n+1)(n!))^2}{2(n+1)(2n+1)(n!)^2}\\=\lim_{n\to\infty}\dfrac{(n+1)^2(n!)^2}{2(n+1)(2n+1)(n!)^2}=\lim_{n\to\infty}\dfrac{n+1}{2(2n+1)}=\frac{1}{4}<1$.

Hence the series converges by the Ratio Test.

Answer 3

This can be off-topic; so, forgive me if this is the case.

Consider $$a_n=\frac{(n!)^2}{(2n)!}$$ and use Stirling approximation for $m!$ $$m!\approx\sqrt{2 \pi } e^{-m} m^{m+\frac{1}{2}}$$ So, $$a_n \approx \sqrt{\pi }\, 2^{-2 n} \sqrt{n}$$ and the ratio test $$\frac{a_{n+1}}{a_n}\approx\frac 14 \sqrt{\frac{n+1}{n}}$$ and $$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\frac 14$$ as shown in the rigorous answers.

Answer 4

$$a_n=\frac{(n!)^2}{(2n)!}=\frac{n!}{(2n)(2n-1)\cdots(n+1)}=\frac{n}{2n}\times\frac{n-1}{2n-1}\times\cdots\times\frac{1}{n+1}<\left(\frac{1}{2}\right)^n,$$

therefore $$\sum_{n=1}^\infty a_n < \sum_{n=1}^\infty \left(\frac{1}{2}\right)^n = 1$$.