Evaluating Limits of Sequences Examples 2

# Evaluating Limits of Sequences Examples 2

We will now look at some more examples of evaluating the limit of a sequence. More examples can be found on the Evaluating Limits of Sequences and Evaluating Limits of Sequences Examples 1 pages.

## Example 1

Evaluate the limit $\lim_{n \to \infty} \frac{n!}{n^n}$.

Expanding the factoring and power of the denominator in the sequence, we get that:

(1)
\begin{align} \lim_{n \to \infty} \frac{n!}{n^n} = \lim_{n \to \infty} \frac{1 \cdot 2 \cdot ... \cdot (n-1) \cdot n}{\underbrace{n \cdot n \cdot ... \cdot n}_{\mathrm{n \: times}}} = \lim_{n \to \infty} \frac{1}{n} \cdot \left ( \frac{2 \cdot 3 \cdot ... \cdot (n-1) \cdot n}{\underbrace{n \cdot n \cdot ... \cdot n}_{\mathrm{n-1\: times}}} \right ) \end{align}

Now note that $\left ( \frac{2 \cdot 3 \cdot ... \cdot (n-1) \cdot n}{\underbrace{n \cdot n \cdot ... \cdot n}_{\mathrm{n-1\: times}}} \right )$ is going to be at most $1$ since the numerator is larger than the denominator. We know that therefore $0 < \frac{n!}{n^n} ≤ \frac{1}{n}$. But since $\lim_{n \to \infty} \frac{1}{n} = 0$, by the sequence theorem it follows that $\lim_{n \to \infty} \frac{n!}{n^n} = 0$.