<s>[INST] 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?
 [/INST] 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.
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