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?
 A: Let $a_n=\frac{2\cdot 4\cdot\ldots\cdot2n}{n^n}$, then
$$\frac{a_{n+1}}{a_n}=\frac{2(n+1)n^n}{(n+1)^{n+1}}=2\times\left(1-\frac{1}{n+1}\right)^n$$
It follows
$$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=2\times\frac{\lim_{n\to\infty}\left(1-\frac{1}{n+1}\right)^{n+1}}{\lim_{n\to\infty}\left(1-\frac{1}{n+1}\right)}=\frac{2e^{-1}}1=\frac2e<1$$
Thus, from the ratio test, $\sum_{n= 1}^{\infty}a_n$ converges.
A: 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.
A: Note that your sum is of the form $\sum_{n=1}^\infty a_n$, where 
$$
a_n=\frac{2\cdot 4\cdot\dots \cdot 2n}{n^n}=\frac{2^nn!}{n^n}.
$$
Thus 
$$
\lim_{n\to\infty}a_n^{1/n}=\lim_{n\to\infty}\frac{2(n!)^\frac{1}{n}}{n}.
$$
By Stirling's formula, $n!$ is approximated by $\left(\frac{n}{e}\right)^n\sqrt{2\pi n}$. Plugging this in, we get 
$$
\lim_{n\to\infty}a_n^{1/n}=\lim_{n\to\infty}\frac{2\left(\left(\frac{n}{e}\right)^n\sqrt{2\pi n}\right)^{\frac{1}{n}}}{n}=\lim_{n\to\infty}\frac{2n}{en}=\frac{2}{e}<1
$$
A: You could also use the Stirling formula in
$$
a_n=\frac{2^n·n!}{n^n}=\frac{2^n·\sqrt{2\pi(n+\theta_n)}\left(\frac ne\right)^n}{n!}\sim \sqrt{n}\left(\frac 2e\right)^n
$$
Then the root test has the limit $\frac 2e<1$ as in the ratio test
A: 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.
