Using D'Alembert's ratio test, determine the convergence
Using D'Alembert's ratio test, determine the convergence of: $$ \sum_{n=1}^{\infty} \frac{n!}{n^n}. $$
Explanation of Method
The ratio $\frac{a_{n+1}}{a_n}=\left(\frac n{n+1}\right)^n=\dfrac1{(1+1/n)^n}$ converges to $\frac1e$ by the standard exponential limit $(1+1/n)^n\to e$. Since $\frac1e\approx0.368<1$, the series converges — this is the reciprocal counterpart to Problem 49's divergent $\sum n^n/n!$, since $n^n$ grows faster than $n!$.
- Compute the ratio $a_{n+1}/a_n$, simplifying to a power of $\frac n{n+1}$.
- Evaluate the limit using $\lim(1+1/n)^n=e$.
- Apply the ratio test to conclude convergence.
Step 1 · Compute the ratio
With $a_n=\dfrac{n!}{n^n}$:
$$ \frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!} = \frac{(n+1)n^n}{(n+1)^{n+1}} = \frac{n^n}{(n+1)^n} = \left(\frac n{n+1}\right)^n. $$
Step 2 · Evaluate the limit
$$ \left(\frac n{n+1}\right)^n = \frac1{\left(1+\frac1n\right)^n} \longrightarrow \frac1e, $$
using the standard limit $\lim_{n\to\infty}\left(1+\frac1n\right)^n=e$.
Step 3 · Apply the ratio test
Since $L=\dfrac1e\approx0.368<1$, by D'Alembert's ratio test $\sum_{n=1}^{\infty}\dfrac{n!}{n^n}$ converges.
Summary Table
| Quantity | Value |
|---|---|
| Ratio $a_{n+1}/a_n$ | $\left(\dfrac n{n+1}\right)^n$ |
| $L=\lim a_{n+1}/a_n$ | $\dfrac1e\approx0.368$ |
| Conclusion | Converges |
Additional Notes
- Stirling's formula gives an independent confirmation: $\dfrac{n!}{n^n}\approx\sqrt{2\pi n}\,e^{-n}$, and this exponential decay $e^{-n}$ easily dominates the polynomial growth of $\sqrt{2\pi n}$, matching the ratio test's conclusion.
- This is precisely the reciprocal series of Problem 49 ($\sum n^n/n!$, which diverges): comparing the two shows that whichever of $n!$ or $n^n$ sits in the numerator determines convergence, since $n^n$ grows strictly faster than $n!$ as $n\to\infty$.
- Successive ratios approach $1/e\approx0.368$ from above ($\frac{a_2}{a_1}=0.5$, $\frac{a_3}{a_2}\approx0.444$, $\frac{a_4}{a_3}\approx0.422$, ...), illustrating the slow but steady convergence of $(1+1/n)^n$ to $e$.
Result
The series converges: