Using Stirling's formula n! = sqrt(2 pi n) n^ne^(-n+
Using Stirling's formula $n! = \sqrt{2\pi n}\,n^ne^{-n+\frac{\theta_n}{12n}}$ ($0<\theta_n<1$), determine the convergence of: $$ \sum_{n=1}^{\infty} \frac{n^n}{n!}. $$
Explanation of Method
Substituting Stirling's formula directly into $n^n/n!$ makes the $n^n$ factors cancel, leaving $\frac{e^n}{\sqrt{2\pi n}}$ (times a correction factor tending to $1$). Since $e^n$ grows exponentially while $\sqrt{2\pi n}$ only grows polynomially, the general term diverges to $\infty$ rather than vanishing, so the Divergence Test settles the matter immediately — and the ratio test confirms it independently.
- Substitute Stirling's formula and simplify.
- Show the general term diverges to $\infty$, triggering the Divergence Test.
- Confirm independently using the ratio test.
Step 1 · Substitute Stirling's formula
$$ a_n = \frac{n^n}{n!} = \frac{n^n}{\sqrt{2\pi n}\,n^ne^{-n+\theta_n/(12n)}} = \frac{e^n}{\sqrt{2\pi n}}\cdot e^{-\theta_n/(12n)}. $$
Since $0<\theta_n<1$, the correction factor $e^{-\theta_n/(12n)}\to1$, so $a_n\sim\dfrac{e^n}{\sqrt{2\pi n}}$.
Step 2 · Apply the divergence test
The exponential $e^n$ dominates the polynomial-rate decay of $\frac1{\sqrt{2\pi n}}$, so $a_n\to\infty$. Since $\lim_{n\to\infty}a_n=\infty\ne0$, by the Divergence Test $\sum_{n=1}^{\infty}\frac{n^n}{n!}$ diverges.
Step 3 · Confirm via the ratio test
$$ \frac{a_{n+1}}{a_n} = \frac{(n+1)^{n+1}/(n+1)!}{n^n/n!} = \frac{(n+1)^{n+1}}{n^n(n+1)} = \left(\frac{n+1}n\right)^n = \left(1+\frac1n\right)^n \to e \approx2.718 > 1. $$
Since the ratio limit exceeds $1$, the ratio test independently confirms divergence.
Summary Table
| Test | Result |
|---|---|
| Divergence test | $a_n\sim e^n/\sqrt{2\pi n}\to\infty$ |
| Ratio test | $a_{n+1}/a_n\to e>1$ |
| Conclusion | Diverges |
Additional Notes
- The terms grow explosively: $a_1=1$, $a_2=2$, $a_3=4.5$, $a_4\approx10.67$, $a_5\approx26.04$ — each roughly $e\approx2.718$ times the previous, exactly matching the ratio test's prediction.
- Intuitively, $n^n/n! = \frac{n\cdot n\cdots n}{1\cdot2\cdots n}$ compares $n$ copies of the largest factor $n$ against the factorial's factors, which average out much smaller (closer to $n/e$ by Stirling), so the numerator systematically outpaces the denominator.
- This problem is a direct converse of Problems 16 and 50, which examine $n/\sqrt[n]{n!}$ and show it tends to the same constant $e$ — the underlying Stirling asymptotic $(n!)^{1/n}\sim n/e$ is the common thread linking all three results.
Result
The series diverges: