Convergence of the sum $\sum_1^{\infty}\frac{(2n)!a^n}{n^{2n}}$

Question

I want to see if the sum $$\sum_1^{\infty}\frac{(2n)!a^n}{n^{2n}}$$ converges or diverges for $a>0$. I know I have to apply the ratio test. Using the ratio test I get that if the limit $$\lim_{n\to\infty}\frac{(2n+1)(2n+2)an^{2n}}{(n+1)^{2n+2}}$$ is $<1$ then it converges. I know the solution, and it will converge if $a<\frac{e^2}{4}$. How do I get to this solution from my limit? I don't know where the $e$ comes from.

Answer 1

Use the fact that$$\lim_{n\to\infty}\frac{(2n+1)(2n+2)}{(n+1)^2}=4$$and that$$\lim_{n\to\infty}\frac{n^{2n}}{(n+1)^{2n}}=\left(\lim_{n\to\infty}\frac1{\left(1+\frac1n\right)^n}\right)^2=\frac1{e^2}.$$