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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