Check if a given series is convergent

Question

I wanted to check if the sum $\sum_{n>0}\frac{n!}{n^n}$ converges.

As $n!$ grows faster than $n^n$, I first thought that it would be divergent. But this doesn't seem to be true, as I found out later using wolfram alpha. So the first thing I don't understand, is why it converges, as normally a factorial function grows faster than a polynomial one?

I then tried to use the root test criteria to check the series:

$lim_{n->\infty} \sqrt[n]{|a_n|}=\sqrt[n]{|\frac{n!}{n^n}|}= |\frac{\sqrt[n]n!}{n}| = |\frac{(n(n-1)!)^\frac{1}{n}}{n}|= |\frac{n^{-n}(n-1)!^\frac{1}{n}}{n}| = |\frac{(n-1)!^{-n}}{n^{n+1}}| $

So I think that the numerator goes to $- \infty$ and the denominator goes to $+\infty$, which leads to an undefined form. Therefore I want to differentiate the numerator and denumerator to apply the rule of de l'Hospital, but again I get an undefined form.

Therefore I tried to solve it using another criteria and started from the beginning:

$lim_{n->\infty} |\frac{a_{n+1}}{a_n}|=|\frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}}=|\frac{n^n{(n+1)!}}{n!(n+1)^{n+1}}|= |\frac{n^n{(n+1)}}{(n+1)^{n+1}}| = |n^n{(n+1)(n+1)^{-n-1}}| = |n^n(n+1)^{-n}|= |\frac{n^n}{n^n+1}|= |\frac{1}{1+\frac{1}{n^n}}|=1 $

As the value is exactly $1$, I can't say if it converges or not. So what am I doing wrong again. Can it in general be solved with one of these two criteria, and what would be the easiest way to solve it?

Answer 1

You are on the right track. You've just misused the ratio test criterion: \begin{align*} \lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_{n}}\right| = \lim_{n\rightarrow\infty}\left(\frac{n}{n+1}\right)^{n} = \lim_{n\rightarrow\infty}\left(1-\frac{1}{n+1}\right)^{n+1} = \frac{1}{e} < 1 \end{align*}

EDIT

Observe that \begin{align*} \frac{n}{n+1} &= \frac{(n+1)-1}{n+1} = 1 - \frac{1}{n+1} \Longrightarrow\\ \left(\frac{n}{n+1}\right)^{n} & = \left(1 - \frac{1}{n+1}\right)^{n} = \left(1 - \frac{1}{n+1}\right)^{n+1}\left(1 - \frac{1}{n+1}\right)^{-1}\Longrightarrow\\ \lim_{n\rightarrow\infty}\left(\frac{n}{n+1}\right)^{n} & = \lim_{n\rightarrow\infty}\left(1-\frac{1}{n+1}\right)^{n+1}\times\lim_{n\rightarrow\infty}\left(1 - \frac{1}{n+1}\right)^{-1} \end{align*} Given that \begin{align*} \lim_{n\rightarrow\infty}\left(1 - \frac{1}{n+1}\right)^{-1} = 1\quad\text{and}\quad\lim_{n\rightarrow\infty}\left(1 + \frac{1}{n}\right)^{n} = e \end{align*} Can you deduce from here the remaining steps?

Answer 2

You made an error in the calculation for the ratio test... (there's an extra factor of $(n+1))$: the limit is in fact $\frac1e\lt1$. So it converges.

(Recall: $\lim_{n\to\infty}(1+\frac1n)^n=e$).

Answer 3

In order to prove the convergence of the given series you only need very crude bounds, like the one provided by the AM-GM inequality, for instance. For any $n>1$ we have $$ \frac{n!}{n^n}=\text{GM}\left(\frac{1}{n},\frac{2}{n},\ldots,\frac{n}{n}\right)^n<\text{AM}\left(\frac{1}{n},\frac{2}{n},\ldots,\frac{n}{n}\right)^n=\left(\frac{n+1}{2n}\right)^n\leq\left(\frac{3}{4}\right)^n $$ hence convergence is granted by a comparison with a geometric series.

Answer 4

For $n\ge 2,$

$$\frac{n!}{n^n}=\frac{n}{n}\frac{n-1}{n}\cdots \frac{2}{n}\frac{1}{n}\le \frac{2}{n^2}.$$

Since $\sum \dfrac{2}{n^2} <\infty,$ $\sum \dfrac{n!}{n^n}<\infty$ by the comparison test.