Pointwise convergence of a series of functions using the ratio test. I am asked to calculate the pointwise convergence of the series of functions 
$$\sum_{n\geq 0} a_n=\sum_{n\geq 0}\frac{(2n)^{\frac{n+1}{2}}}{\sqrt{n!}}x^ne^{-nx^2}$$
Since it is a series of positive terms, I apply the ratio test and after simplifying $\frac{a_{n+1}}{a_n}$ I got to $$\frac{a_{n+1}}{a_n}=\frac{x\sqrt{2}((1+1/n)^n(1+1/n))^{1/2}}{e^{x^2}}$$
As a hint, I was told that I should use $$(1+1/n)^n\geq\frac{e}{1+1/n}$$ For all $n\geq 1$
Therefore the initial series will converge if $$1>\frac{a_{n+1}}{a_n}\geq \frac{x\sqrt{2e}}{e^{x^2}}$$
How do I finish the problem? If I had $\frac{a_{n+1}}{a_n}= \frac{x\sqrt{2e}}{e^{x^2}}$ I would say it converges for all $x$ such that $e^{x^2}<x\sqrt{2e}$.
 A: By ratio test we obtain
$$\left|\frac{[2(n+1)]^{\frac{n+2}{2}}}{\sqrt{(n+1)!}}\frac{\sqrt{n!}}{[2n]^{\frac{n+1}{2}}}\frac{x^{n+1}e^{-(n+1)x^2}}{x^ne^{-nx^2}}\right|=\sqrt 2|x|e^{-x^2}\left(1+\frac{1}{n}\right)^\frac{n+1}2\to \sqrt {2e}|x|e^{-x^2}\le 1$$
for $\sqrt {2e}|x|e^{-x^2}=1 \implies |x|e^{-x^2}=\frac1{\sqrt{2e}} $ we have 
$$\left|\frac{2n^{\frac{n+1}{2}}}{\sqrt{n!}}x^ne^{-nx^2}\right|=\frac{(2n)^{\frac{n+1}{2}}}{\sqrt{n!}}\frac1{(2e)^\frac n2}\sim \frac{(2n)^{\frac{n+1}{2}}}{\sqrt[4]{2\pi n}n^\frac n 2}\frac{e^\frac n 2}{(2e)^\frac n2}=\frac{\sqrt{2n}}{\sqrt[4]{2\pi n}} \to \infty$$
Therefore the given series converges for any $x\in \mathbb R$ such that  $\sqrt {2e}|x|e^{-x^2}<1$.
A: This is just a supplement to user's excellent answer, showing how the hint was supposed to be applied.  As shown in that answer, the ratio $$\sqrt 2|x|e^{-x^2}\left(1+\frac{1}{n}\right)^{(n+1)/2}\to r,$$ where $r<1$ unless $\sqrt 2|x|e^{-x^2}=1$.  In that case, the ratio is $$\left(1+\frac{1}{n}\right)^{(n+1)/2}>\sqrt{e\over1+1/n}\to\sqrt e>1,$$ by the hint.
