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How do you use the ratio test to show whether

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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.)

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

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