<s>[INST] 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
 [/INST] 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 $\sum_{n \geq 1} \binom{2n}{n} \frac{1}{4^n}$.
</s>