Test the absolute and conditional convergence
Test the absolute and conditional convergence of the series: $$ \sum_{n=1}^{\infty}(-1)^n \frac{n!}{n^n}. $$
Explanation of Method
The ratio test applied directly to the absolute series $\sum\frac{n!}{n^n}$ produces the classical limit $1/e$, since the ratio simplifies to $\left(\frac n{n+1}\right)^n\to\frac1e$. Since $1/e<1$, absolute convergence follows immediately — no need for Leibniz's test or any finer analysis.
- Apply the ratio test to $\sum\frac{n!}{n^n}$.
- Simplify the ratio to $\left(\frac n{n+1}\right)^n$ and evaluate its limit.
- Conclude absolute convergence, verified independently via Stirling's approximation.
Step 1 · Set up the ratio test
With $|a_n|=\dfrac{n!}{n^n}$:
$$ \frac{|a_{n+1}|}{|a_n|} = \frac{(n+1)!\,n^n}{n!\,(n+1)^{n+1}} = \frac{n^n}{(n+1)^n} = \left(\frac n{n+1}\right)^n. $$
Step 2 · Evaluate the limit
$$ \lim_{n\to\infty}\left(\frac n{n+1}\right)^n = \lim_{n\to\infty}\frac1{\left(1+\frac1n\right)^n} = \frac1e. $$
Step 3 · Apply the ratio test and verify via Stirling
Since $\frac1e<1$, by the ratio test $\sum\frac{n!}{n^n}$ converges, so the original alternating series converges absolutely. Confirming via Stirling's formula: $\dfrac{n!}{n^n}\sim\dfrac{\sqrt{2\pi n}}{e^n}$, and $\sum\dfrac{\sqrt n}{e^n}$ converges easily by the ratio test (limit $1/e<1$ again), consistent with the direct computation.
Summary Table
| Quantity | Value |
|---|---|
| $\lim|a_{n+1}/a_n|$ | $1/e\approx0.368$ |
| Stirling asymptotic | $n!/n^n\sim\sqrt{2\pi n}/e^n$ |
| Conclusion | Absolutely convergent |
Additional Notes
- The limit $1/e$ arising from $\left(\frac{n}{n+1}\right)^n\to1/e$ is the same fundamental constant that appears throughout Stirling's-formula-based problems (Problems 51–66, 89), reflecting the deep connection between factorial growth and the exponential constant.
- Since the series converges absolutely, the alternating sign pattern is essentially irrelevant to convergence — the terms decay so fast (geometrically, with ratio approaching $1/e\approx0.368$) that no finer test like Leibniz's or Dirichlet's is ever needed.
- This is a useful contrast with Problem 89's $\sum\frac1{n!}\left(\frac ne\right)^n$, the reciprocal-style construction: there the ratio test gave exactly $1$ (inconclusive), requiring Raabe's test to resolve; here, inverting the roles of $n!$ and $n^n$ produces a clean ratio strictly less than $1$, making this problem far simpler to resolve.
Result
The series converges absolutely: