Does $\sum_{n=1}^\infty \frac{2\cdot 4\cdot 6\cdot ...\cdot (2n)}{n^n}$ converge? Does $\sum_{n=1}^\infty \frac{2\cdot 4\cdot 6\cdot ...\cdot (2n)}{n^n}$ converges?
Well, I'm trying the approach using Cauchy's test for convergence. If $a_n >= 0$ and $\lim \sqrt[n]a_n = L $ exists. then, I need to check whether it is bigger or lesser than $1$. (if equal then we have a problem)
So.. $\sum_{n=1}^\infty \frac{2\cdot 4\cdot 6\cdot ...\cdot (2n)}{n^n}$ becomes $\sum_{n=1}^\infty \frac{\sqrt[n]{2\cdot 4\cdot 6\cdot ...\cdot (2n)}}{n}$ becomes $\frac{({2\cdot 4\cdot 6\cdot ...\cdot (2n)})^{1/n}}{n}$ becomes computing the limit of $1/n$ which is zero. Which means it converges?

Observe this variant of rational test.
$$\sum_{n=1}^\infty \frac{2\cdot 4\cdot 6\cdot ...\cdot (2n)}{n^n}=\sum_{n=1}^\infty 2^n \frac{n!}{n^n}=\sum_{n=1}^\infty 2^n \prod\limits_{k=1}^{n} \frac{k}{n}$$
Now $$\sum_{n=1}^\infty 2^n \cdot b_{n}$$ could be understood as sort of a binary expansion and we are sure that it converges as long as $$b_{n} < \frac{k}{d^{n}}, \, \, \, d > 2$$ for all $n>M$ for some fixed $k$ and $M$.
This is the same as requiring
$$\frac{b_{n+1}}{b_{n}} < \frac{1}{d}$$
for some $n>M$ but indeed we have
$$\frac{\prod\limits_{k=1}^{n+1} \frac{k}{n+1}}{\prod\limits_{k=1}^{n} \frac{k}{n}}=(\frac{1}{1+\frac{1}{n}})^{n}$$
Since $\lim\limits_{n \to \infty}(\frac{1}{1+\frac{1}{n}})^{n}=\frac{1}{e}$ we are certain that there are $M$ and $d$ for which $$\frac{b_{n+1}}{b_{n}} < \frac{1}{e} < \frac{1}{d} < \frac{1}{2} $$
For this reason the series converges.
