Convergence of $\sum \frac{(2n)!}{n!n!}\frac{1}{4^n}$ Does the series
$$\sum \frac{(2n)!}{n!n!}\frac{1}{4^n}$$
converges?
My attempt: Since the ratio test is inconclusive, my idea is to use the Stirling Approximation for n!
$$\frac{(2n)!}{n!n!4^n} \sim (\frac{1}{4^n} \frac{\sqrt{4\pi n}(\frac{2n}{e})^{2n}}{\sqrt{2 n \pi} \sqrt{2n \pi} (\frac{n}{e})^{2n}} =\frac{(2)^{2n}}{4^n \sqrt{n \pi}}$$
The series of the secomd term diverges. It is correct to conclude thatthe series diverges?
Another ideas are welcome!
Thanks
 A: 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.
A: There is an identity: $$\sum_{k=0}^n {{2k}\choose k} {{2n-2k}\choose{n-k}} = 4^n$$
From this, we can calculate, for all $x$ with $|x|<1/4$:
$$\left(\sum_{n=0}^\infty {{2n}\choose n} x^n\right)^2 = \sum_{n=0}^\infty \left( \sum_{k=0}^n {{2k}\choose k} {{2n-2k}\choose{n-k}}\right)x^n = \sum_{n=0}^\infty 4^n x^n = \frac{1}{1-4x}$$
We conclude:
$$\sum_{n=0}^\infty {{2n}\choose n} 4^{-n} = \lim_{x\to 1/4} \sum_{n=0}^\infty {{2n}\choose n} x^n = \lim_{x\to 1/4^-} \frac{1}{\sqrt{1-4x}} = \infty$$
A: 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 $\frac cn$.


*

*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.

*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.

*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
$$ \int_0^{2\pi} \cos^{2n} t \,dt
= 4\int_0^{\pi/2} \cos^{2n} t \,dt
\ge 4\int_0^{\pi/2} \cos^{2n} t \sin t \,dt
= \frac4{2n+1}
$$

A: 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 $\sum_{n \geq 1} \binom{2n}{n} \frac{1}{4^n}$.
A: In this answer it is shown that
$$
\frac{4^n}{\sqrt{\pi(n+\frac13)}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi(n+\frac14)}}
$$
Therefore,
$$
\begin{align}
\frac{(2n)!}{n!\,n!}\frac1{4^n}
&\ge\frac1{\sqrt{\pi(n+\frac13)}}\\
&\ge\frac1{\pi(n+1)}\\
\end{align}
$$
compare with the Harmonic Series.
