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

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?

• The Ratio Test works here, you just did not compute the limit correctly. – André Nicolas Jun 17 '13 at 19:41
• The problem you made for yourself is that your denominator should contain $\ (n+1)^{n+1} \$ . – colormegone Jun 17 '13 at 19:43

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

• I'm sorry, but I'm not following. All I learned so far was to do algebraic manipulation of the ratio $a_{n+1}/a_n$ and then find the subsequent limit. – Mirrana Jun 17 '13 at 19:39
• Oh, I see now - I messed up with the $a_{n+1}$ term. I'll look at it again, thanks. – Mirrana Jun 17 '13 at 19:45

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

• That would be it. Thanks, I couldn't see that when I looked over it. – Mirrana Jun 17 '13 at 19:45

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.

• How did you get from $n^n/(n+1)^n$ to $1/(1+1/n)^n$? I can understand the rest after that.. – Mirrana Jun 17 '13 at 22:24
• @agent154: divide both numerator and denominator by $n^n$. – robjohn Jun 17 '13 at 23:38
• That's probably a whole lot better than the way I did it by using L'Hopital's rule all over again with $(1-1/(n+1))^n$ – Mirrana Jun 18 '13 at 0:11