# This is a ratio test question.

How do you use the ratio test to show whether

converges or diverges?

Using symbolab gave me something with "series root test", however it is not covered in my course. Would it be possible to show divergence/convergence without this? Thanks.

(The answer, according to the solutions, is converges.)

• Let $a_n=\frac{n^{2n}}{3^n(2n)!}$. You must compute $\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|$.
– mjh
May 30 '14 at 15:15

Let $a_{n}$ be the $n$-th term of our series. The ratio $\frac{a_{n+1}}{a_n}$ easily simplifies to $$\frac{(n+1)^{2n+2}}{3n^{2n}(2n+2)(2n+1)}.$$ Writing $(n+1)^{2n+2}$ as $(n+1)^2(n+1)^{2n}$ yields the further simplification
$$\frac{n+1}{6(2n+1)}\cdot \left(1+\frac{1}{n}\right)^{2n}.$$ The limit as $n\to\infty$ of $\frac{n+1}{6(2n+1)}$ is $\frac{1}{12}$, while $\left(1+\frac{1}{n}\right)^{2n}$ has limit $e^2$.
It follows that $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\frac{e^2}{12}\lt 1$, so by the Ratio Test our series converges.