Does the series $\sum\frac{(n!)^2\cdot4^n}{(2n)!}$ diverge or converge? Does the series $\sum \frac{(n!)^2\cdot4^n}{(2n)!}$ converge or diverge?
 A: $4^n = \sum_{k=0}^{2n} \binom{2n}{k} \ge \binom{2n}{n}$, so the terms of this series are all at least one.  (In particular, they don't converge to zero.)
A: $\dfrac{(n!)^24^n}{(2n)!}=\dfrac{2^n\prod_{r=1}^n(2r)}{2^n\prod_{r=1}^n(2r-1)}=\dfrac{\prod_{r=1}^n(2r)}{\prod_{r=1}^n(2r-1)}$
Now $\dfrac{2r}{2r-1}>1$ for $2r-1>0$
A: A bit useless in light of Steven's answer, but nonetheless useful for future ideas. Recall that the Catalan numbers are equal to $\frac{1}{n+1}\binom {2n}n$ and count the number of sequences of length $2n$ of zeros and ones for which the number of ones is always at least that of zeros at any truncation of the sequence. Since there are $4^n$ binary sequences of length $2n$ it follows that $$\frac{1}{n+1}\leqslant 4^n \binom{2n}n^{-1}$$ so the series diverges by comparison to the harmonic series. In fact $$\frac{1}{4^n}\binom{2n}n \simeq \frac{1}{\sqrt{n\pi }}$$
A: Recall Stirling's Formula 
$$n!=\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(1+O\left(\frac1n\right)\right)$$
Then, we have
$$\begin{align}
\frac{(n!)^2\,4^n}{(2n)!}&=\frac{(2\pi n)\left(\frac{n}{e}\right)^{2n}4^n}{\sqrt{4\pi n}\left(\frac{2n}{e}\right)^{2n}}\left(1+O\left(\frac1n\right)\right)\\\\
&=\sqrt{\pi n}+O\left(\frac{1}{\sqrt{n}}\right)\\\\
&\to \infty\,\,\text{as}\,\,n\to \infty
\end{align}$$
