## Calculus 8th Edition

Use the Ratio Test: $\lim_\limits{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|$ Let $$a_n=\frac{n!}{{e^n}^2}$$ Then $$\lim_\limits{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_\limits{n\to\infty}\left|\frac{(n+1)!}{{e^{(n+1)}}^2}\cdot\frac{{e^n}^2}{n!}\right|$$ We can take away the absolute value bars and then simplify. $$\lim_\limits{n\to\infty}\frac{(n+1)n!\cdot {e^n}^2}{e^{n^2+2n+1}n!}=\lim_\limits{n\to\infty}\frac{(n+1)}{e^{2n+1}}$$ so, as $n\to \infty$, $$\lim_\limits{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|\to0$$ thus, the series is convergent by the Ratio Test.