Worked exercise — The sequence n! / n^n

Problem

Show that the sequence $a_n = \dfrac{n!}{n^n}$ is decreasing and compute its limit.

Solution

We study the ratio of two consecutive terms: $\frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!} = \frac{(n+1)\,n^n}{(n+1)^{n+1}} = \frac{n^n}{(n+1)^n} = \left(\frac{n}{n+1}\right)^n = \frac{1}{\left(1+\frac1n\right)^n}.$ Since $\left(1+\frac1n\right)^n \to e > 1$, the ratio tends to $\dfrac{1}{e} < 1$; in particular $\dfrac{a_{n+1}}{a_n} < 1$ for every $n$, so the sequence is decreasing (and it is positive, hence bounded below by $0$).

Being decreasing and bounded below, it converges; and since the ratio of consecutive terms tends to $1/e < 1$, the terms decrease geometrically towards $0$: $\lim_{n\to+\infty}\frac{n!}{n^n} = \boxed{0}.$

Links

Topics: Sequences
Concepts: Limit of a sequence · Monotonicity · Euler’s number
Skills: Computing limits · Proving
Exercise type: Computing a limit · Proof