Checking if a series $\sum_{n=1}^\infty \frac{(2n)!}{(n!)^2}$ is divergent or not The series $$\sum_{n=1}^\infty \frac{(2n)!}{(n!)^2}$$ 
Is divergent.
Is this statement true?
I tried by Root Test.
 A: The terms $\binom{2n}{n}= \frac{(2n)!}{(n!)^2}$ in this summation are called central binomial coefficients because in each row of Pascal's triangle, the middle entry is the largest.
Rather than approaching zero (as convergence of a series requires), these terms grow roughly like $4^n$.  A more precise estimate is given by the Wikipedia article (linked above) and proved to be an upper bound in this previous MathOverflow Question:
$$ \binom{2n}{n} \approx \frac{4^n}{\sqrt{\pi n}} $$
Indeed the related power series $\sum_{n=0}^\infty \binom{2n}{n} x^n$ converges to $\frac{1}{\sqrt{1-4x}}$ for $x\in [-1/4,1/4)$.  One might suspect that the original form of the problem that inspired this Question was to determine the radius of convergence of this power series (which is $R=1/4$ by location of the nearest singularity to the origin).
A: Yes, this series is clearly divergent because each term it does not approach $0$ as $n$ approaches $\infty$. If we split up
$$\frac{(2n)!}{(n!)^2}$$
like this:
$$\frac{n!\cdot(n+1)(n+2)(n+3)...(2n)}{n!n!}$$
$$\frac{\cdot(n+1)(n+2)(n+3)...(2n)}{n!}$$
$$\frac{n+1}{1}\cdot\frac{n+2}{2}\cdot...\cdot\frac{n+n}{n}$$
Then it becomes apparent that each term is always greater than one, and thus the sum obviously diverges.
A: $$\frac{(2n)!}{n!^2}=\frac{1\cdot 2\cdot\ldots \cdot n\,\cdot\,(n+1)\cdot\ldots \cdot 2n}{1\cdot 2\cdot\ldots \cdot n\,\cdot\,1\cdot 2\cdot\ldots \cdot n}$$
After cancellatoin, the numerator consists of $n$ factors $\ge n$, the denominator of $n$ factors $\le n$. Hence the general summand is $\ge 1$, but miust tend to $0$ for convergence.
