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?

Note that $\frac{2*4*6*...*2n}{n^n}=\frac{n!2^n}{n^n}$.
By the ratio test, $\dfrac{\left(\frac{2}{n+1}\right)^{n+1}(n+1)!}{\left(\frac{2}{n}\right)^{n}(n)!}=\dfrac{2n^n}{(n+1)^n}=2\left(\dfrac{n}{n+1}\right)^n=2\left(1-\frac{1}{n+1}\right)^n\to\frac{2}{e}<1$.
Thus the serie is convergent.
