Question

In this task, use the Ratio Test to determine the convergence or divergence of the series.

$\displaystyle\sum_{n=1}^{\infty} \frac{n^n}{n !}$

Solution

VerifiedAnswered 2 years ago

Answered 2 years ago

Let $\sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}\dfrac{n^n}{n!}$. Consider

$\begin{aligned}\lim_{n\rightarrow \infty}\left|\dfrac{a_{n+1}}{a_n}\right|&=\lim_{n\rightarrow \infty}\left[\dfrac{(n+1)^{n+1}}{(n+1)!}\cdot \dfrac{n!}{n^n}\right]\\&=\lim_{n\rightarrow \infty}\dfrac{(n+1)^{n+1}}{n^n(n+1)}\\&=\lim_{n\rightarrow \infty}\dfrac{(n+1)^n}{n^n}\\&=\lim_{n\rightarrow \infty}\left(1+\dfrac{1}{n}\right)^n\\&=e\\&>1\end{aligned}$

Thus using ratio test we get $\sum_{n=1}^{\infty}a_n$ diverges.

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