Convergence of $\sum \frac{(2n)!}{n!n!}\frac{1}{4^n}$

Solution 1:

Another approach: Note that $\frac{(2n)!}{n!n!}$ equals the binomial coefficient $\binom{2n}{n}$. The binomial coefficients $\binom{2n}{0}$, $\binom{2n}{1}$, ..., $\binom{2n}{2n}$ have sum $2^{2n} = 4^n$, thus their average is $\frac{4^n}{2n+1}$. Since $\binom{2n}{n}$ is the largest of these binomial coefficients, it is surely "above average": this means that $\binom{2n}{n} \geq \frac{4^n}{2n+1}$ from which it follows that $$\frac{(2n)!}{n!n!} \frac{1}{4^n} = \binom{2n}{n} \frac{1}{4^n} \geq \frac{1}{2n+1}.$$ Since the series $\sum_{n \geq 1} \frac{1}{2n+1}$ diverges, the same holds for the series $\sum_{n \geq 1} \binom{2n}{n} \frac{1}{4^n}$.

Solution 2:

A much simpler way: $$ \frac{a_{n+1}}{a_n}=\frac{(2n+2)(2n+1)}{4(n+1)(n+1)}=\frac{2n+1}{2n+2}\ge \sqrt{\frac{n}{n+1}}, $$ since $$ \left(\frac{2n+1}{2n+2}\right)^2=\left(1-\frac{1}{2n+2}\right)^2\ge 1-\frac{2}{2n+2} =\frac{n}{n+1}, $$ and hence $$ a_n=a_1\prod_{k=2}^n\frac{a_{k}}{a_{k-1}}\ge a_1\prod_{k=2}^n\sqrt{\frac{k-1}{k}} =\frac{a_1}{\sqrt{n}}, $$ and hence $$ \sum_{n=1}^\infty a_n=\infty. $$

Note that $$ \frac{a_{n+1}}{a_n}=\frac{(2n+2)(2n+1)}{4(n+1)(n+1)}=\frac{2n+1}{2n+2}=1-\frac{1}{2n+2}=1-\frac{1}{2n}+{\mathcal O}\left(\frac{1}{n^2}\right). $$ This series diverges due to Gauss Test otherwise known as Raabe's Test.

Solution 3:

As you noted, Stirling gives an estimate $4^{-n}\binom{2n}{n} \ge \frac{c}{\sqrt n}$, which shows divergence; as others have said, it's enough to show $4^{-n}\binom{2n}{n} \ge \frac cn$, which is much easier than the Stirling estimate. Here are three more arguments for $\frac cn$.

1. The most elementary one that I know (except perhaps the answer by user133281): $$ 4^n = \left(\sum_{k=0}^n 1\cdot\binom nk\right)^2 \le \left(\sum_{k=0}^n 1^2\right) \left(\sum_{k=0}^n \binom nk^2\right) = (n+1)\binom{2n}{n} $$ That's Cauchy-Schwarz and the identity proved in this question.
2. Using the beta function representation of binomial coefficients: $$ \frac1{(2n+1)\binom{2n}{n}} = \int_0^1 t^n (1-t)^n \,dt \le \frac1{4^n} $$ since $\sqrt{t(1-t)}\le\frac12$ by AM/GM.
3. Using the Wallis integral representation of central binomial coefficients, that is, $$ \binom{2n}{n} = \frac{4^n}{2\pi} \int_0^{2\pi} \cos^{2n} t \,dt $$ (You can prove this by induction; a slicker way is to recall that, for $n\in\mathbb Z$, $$\frac1{2\pi} \int_0^{2\pi} e^{int} \,dt = \begin{cases} 1 &\text{if $n=0$} \\ 0 &\text{otherwise} \end{cases}$$ Consequently, if $f(x) = \sum_{k=m}^n a_k x^k$ is a Laurent polynomial then $\frac1{2\pi} \int_0^{2\pi} f(e^{it}) \,dt = a_0$. Now take $f(x)=(x+x^{-1})^{2n}$.) Perhaps the simplest way to estimate the Wallis integral is $$ \int_0^{2\pi} \cos^{2n} t \,dt = 4\int_0^{\pi/2} \cos^{2n} t \,dt \ge 4\int_0^{\pi/2} \cos^{2n} t \sin t \,dt = \frac4{2n+1} $$