# Interesting limit from real analysis: lim n!/n^n

In my Advanced Calculus (introductory real analysis) course at Western Michigan University, Dr. Ledyaev gave us this limit as a bonus homework problem to turn in:

\begin{align} \lim_{n \to \infty}\frac{n!}{n^n} = ~? \end{align}

The answer is that the sequence converges and its limit is 0. Here is how I showed this:

Claim: The sequence $$\{a_n: a_n = \frac{n!}{n^n}\}$$ is monotonically decreasing and bounded below.

Proof: Note that all terms of both the numerator $$n!$$ and the denominator $$n^n$$ are positive for all $$n \in \mathbb{N}$$, so $$\{a_n\}$$ is bounded below (by $$0$$). To determine whether the sequence is increasing or decreasing, we can use the ratio test:
$$\begin{eqnarray} \frac{a_{n+1}}{a_n} &=& \frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}} \\[3pt] \nonumber \\ &=& \frac{(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{n!} \\[3pt] \nonumber \\ &=& \frac{n!(n+1)}{(n+1)^{n+1}} \cdot \frac{n^n}{n!} \\[3pt] \nonumber \\ &=& \frac{n+1}{(n+1)(n+1)^n} \cdot n^n \\[3pt] \nonumber \\ &=& \frac{1}{(n+1)^n} \cdot n^n \\[3pt] \nonumber \\ &=& \frac{n^n}{(n+1)^n} < 1 \nonumber \\ \end{eqnarray}$$ The ratio $$\frac{a_{n+1}}{a_n} < 1$$ for all $$n$$, meaning the sequence is monotonically decreasing. Since it is bounded below and monotonically decreasing, it converges to a limit. $$\tag*{\blacksquare}$$ Claim: $$\lim_{n\rightarrow\infty} \frac{n!}{n^n} = 0.$$

Proof: Since $$\{a_n\}$$ converges to a limit, call the limit $$L$$. An important theorem states that $$\{a_{n+1}\}$$ also converges to $$L$$. From above,
$$\begin{eqnarray} \frac{a_{n+1}}{a_n} &=& \frac{n^n}{(n+1)^n} \\[3pt] \nonumber \\ {a_{n+1}} &=& \frac{n^n}{(n+1)^n} \cdot {a_n} \\[3pt] \nonumber \\ \end{eqnarray}$$

And using the product rule for limits,
$$\begin{eqnarray} \lim_{n \to \infty}{a_{n+1}} &=& \lim_{n \to \infty}\frac{n^n}{(n+1)^n} \cdot \lim_{n \to \infty}{a_n} \\[3pt] \nonumber \\ L &=& \lim_{n \to \infty}\frac{n^n}{(n+1)^n} \cdot L \nonumber \\ \end{eqnarray}$$

If we can show that $$\frac{n^n}{(n+1)^n}$$ converges to some real number $$r$$, then we will have $$L = r \cdot L$$. If $$r \neq 1$$, then the only solution to that equation is $$L = 0$$. In fact, we will show that $$\frac{n^n}{(n+1)^n}$$ converges to $$\frac{1}{e}$$. Observe,

$$\begin{eqnarray} \frac{n^n}{(n+1)^n} = \left(\frac{n}{n+1}\right)^n = \left(\frac{1}{\frac{n+1}{n}}\right)^n = \left(\frac{1}{1+\frac{1}{n}}\right)^n = \frac{1}{\left(1+\frac{1}{n}\right)^n} \end{eqnarray}$$

Recall that $$\lim_{}(1+\frac{1}{n})^n = e$$, so by the quotient rule for limits, $$\lim{}\frac{1}{(1+\frac{1}{n})^n} = \frac{1}{e}$$. Substituting $$\frac{1}{e}$$ for $$r$$ above, we have $$L = \frac{1}{e} \cdot L$$, so $$L = \lim_{}a_n = 0$$. $$\tag*{\blacksquare}$$

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