# Determining the value to which the sequence $a_n=\frac{n!}{n^n}$ converges.

How can it be deduced that the sequence $a_n=\dfrac{n!}{n^n}$ converges to $0$? I can reasonably infer this to be true, because I see the pattern as $n$ approaches larger values, but I am unsure of how to to evaluate $\lim a_n=L$ to show that $L=0$.

• Stirling's approximation. – Robert Israel Jul 8 '15 at 2:09
• @RobertIsrael We've never discussed what you speak of in our class. Do you know of a more elementary solution to determining this limit? – alxmke Jul 8 '15 at 2:11

Write $\frac{n!}{n^n}$ as $$\frac{n}{n}\frac{n-1}{n} \cdots \frac{1}{n}$$ and as you can see by the last factor of this, $\frac{n!}{n^n} \leq \frac1n$ (because all the other factors are less than $1$). Obviously we have $0 \leq \frac{n!}{n^n} \leq \frac 1n$ , so we know that it has to converge to $0$

• Wouldn't I have to squeeze it so that $0 < \frac{n!}{n^n} \leq \frac{1}{n}$ to show that it must converge to 0? – alxmke Jul 8 '15 at 2:18
• yes, but of course $0<\frac{n!}{n^n}$ holds – supinf Jul 8 '15 at 2:25
• In fact, we can find the speed of convergence by looking at the ratio of successive terms. We find that $\frac{a_{n+1}}{a_n}\to e^{-1}$. – Mark Viola Jul 8 '15 at 3:12

For $n > 1$, $$\frac{n!}{n^n} = \prod_\limits{i = 1}^{n}\frac{i}{n} = \frac{1}{n} \prod_\limits{i = 2}^{n}\frac{i}{n} \leq \frac{1}{n}$$ It is bounded by a sequence $\frac{1}{n}$ that converges to zero, so it must converge to zero to.

• $\{\frac1n\}_{n\ge 1}$ is a sequence, not a series! – Mercy King Jul 8 '15 at 2:12
• @Mercy indeed, fixed. – lisyarus Jul 8 '15 at 2:13

HINT:

\begin{align} \frac{a_{n+1}}{a_n}&=\frac{(n+1)!}{(n+1)^{n+1}}\frac{n^n}{n!}\\\\ &=\frac{1}{\left(1+\frac1n\right)^n}\\\\ &\to e^{-1}\\\\ &<1 \end{align}

Of course, Stirling's formula is the better way to go. Although the OP mentioned that that topic has not yet been discussed, I thought that it might be instructive to show how it can be used here. So, here we go ...

Stirling's formula is given by

$$n!=\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(1+O\left(\frac1n\right)\right)$$

Thus, we have

$$\frac{n!}{n^n}=\sqrt{2\pi n}e^{-n}\left(1+O\left(\frac1n\right)\right)$$

which clearly goes to zero as $n \to \infty$.

It is also worth remarking that from Stirling's formula we obtain

$$\frac{a_{n+1}}{a_n}=e^{-1}\left(1+O\left(\frac1n\right)\right)$$

which goes to $e^{-1}$ as $n\to \infty$ as expected.