How to show convergence or divergence of a series when the ratio test is inconclusive?

Question

Use the ratio or the root test to show convergence or divergence of the following series. If inconclusive, use another test:

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

So my first instinct was to try the ratio test due to the existence of the factorial. This is my working:

---

Using the Ratio Test: \begin{align*} \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right|&=\lim_{n\to\infty}\left|\frac{(n+1)\cdot n!\cdot n^{n}}{n!\cdot n\cdot n^{n}}\right|\\ &=\lim_{n\to\infty}\left|\frac{n+1}{n}\right|\\ &=1 \end{align*} The Ratio Test is inconclusive.

---

I decided then to try the root test due to the presence of the $n^n$, but I think that's problematic and won't work (unless I'm looking at something the wrong way). I end up with the following:

\begin{align} \lim_{n\to\infty}\left|a_n\right|^{1/n}&=\lim_{n\to\infty}\left|\frac{n!}{n^n}\right|\\ &=\lim_{n\to\infty}\left|\frac{(n!)^{1/n}}{n}\right|=\frac{\infty}{\infty} \end{align} So my problem here is that I can't apply L'Hopital's rule. If I expand the numerator I get the following: $$(n!)^{1/n}=\sqrt[n]{n}\cdot\sqrt[n]{n-1}\cdot\sqrt[n]{n-2}\cdot\sqrt[n]{n-3}\cdots\sqrt[n]{3}\cdot\sqrt[n]{2}\cdot\sqrt[n]{1}$$

Which then would only allow me to cancel the $\sqrt[n]{n}$ and get $n^{(n-1)/n}$ in the denominator. Still gives me the indeterminate form of $\infty/\infty$.

So how can I approach this? Or was I on the right track and did something wrong?

Answer 1

$$ \frac{n^n}{(n+1)^{n+1}} = \frac{1}{n+1}\cdot\frac{n^n}{(n+1)^n}. $$ The $n+1$ in the denominator is canceled by the $n+1$ in the other part of the whole expression. Then we have $$ \frac{n^n}{(n+1)^n}\to \frac1e\text{ as }n\to\infty. $$

Answer 2

You've got the wrong formula for $a_{n+1}/a_n$. What happened to the $(n+1)^{n+1}$ in $a_{n+1}$?

Answer 3

To make more explicit what Michael Hardy wrote: $$ \begin{align} \left.\frac{(n+1)!}{(n+1)^{n+1}}\middle/\frac{n!}{n^n}\right. &=\frac{(n+1)!}{n!}\frac{n^n}{(n+1)^{n+1}}\\ &=(n+1)\frac{n^n}{(n+1)^n(n+1)}\\ &=\frac1{\left(1+\frac1n\right)^n}\\ &\to\quad\frac1e \end{align} $$ Therefore, this series passes the ratio test and converges.