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: $$ \sum_{n=1}^{\infty} \frac{1}{n!}\left(\frac{n}{e}\right)^n. $$
Explanation of Method
Substituting Stirling's formula into the general term causes the $n^n$ and $e^{-n}$ factors to cancel almost completely against the $(n/e)^n$ factor, leaving only $\frac1{\sqrt{2\pi n}}$ times a correction tending to $1$. Since $n^{-1/2}$ decays too slowly for convergence ($p=\frac12<1$), the limit comparison test with $\sum1/\sqrt n$ shows divergence.
- Substitute Stirling's formula into the general term and simplify.
- Identify the resulting order, $n^{-1/2}$.
- Apply the limit comparison test against $\sum1/\sqrt n$.
Step 1 · Substitute Stirling's formula
$$ a_n = \frac1{n!}\left(\frac ne\right)^n = \frac{n^n}{n!\cdot e^n}. $$
Using $\frac1{n!}=\dfrac{e^n}{\sqrt{2\pi n}\,n^n}\cdot e^{-\theta_n/(12n)}$:
$$ a_n = \frac{n^n}{e^n}\cdot\frac{e^n}{\sqrt{2\pi n}\,n^n}\cdot e^{-\theta_n/(12n)} = \frac1{\sqrt{2\pi n}}\cdot e^{-\theta_n/(12n)}. $$
Step 2 · Identify the order
Since $0<\theta_n<1$, $e^{-\theta_n/(12n)}\to1$, so $a_n\sim\dfrac1{\sqrt{2\pi n}}$ — order $n^{-1/2}$.
Step 3 · Limit comparison test
With $b_n=\dfrac1{\sqrt n}$:
$$ \lim_{n\to\infty}\frac{a_n}{b_n} = \lim_{n\to\infty} \frac{e^{-\theta_n/(12n)}}{\sqrt{2\pi}} = \frac1{\sqrt{2\pi}}. $$
Positive and finite; since $\sum\frac1{\sqrt n}$ diverges ($p=\frac12<1$), by the limit comparison test $\sum_{n=1}^{\infty}\frac1{n!}\left(\frac ne\right)^n$ diverges too.
Summary Table
| Quantity | Value |
|---|---|
| General term $a_n$ | $\sim\dfrac1{\sqrt{2\pi n}}$ |
| Effective exponent | $\dfrac12<1$ |
| Limit comparison ratio (vs. $1/\sqrt n$) | $\dfrac1{\sqrt{2\pi}}$ |
| Conclusion | Diverges |
Additional Notes
- The near-total cancellation in Step 1 is the entire content of Stirling's formula: it says precisely that $n!$ and $\sqrt{2\pi n}\,(n/e)^n$ agree up to a factor tending to $1$, so their ratio's reciprocal leaves only the "leftover" $\sqrt{2\pi n}$ term.
- Numerically this matches Problem 52's own verification: $a_{10}\approx0.126$, and $\frac1{\sqrt{2\pi\cdot10}}\approx0.126$ — an excellent match even at a modest value of $n$.
- This series sits at a specific point along the family $\sum n^p/n!\cdot n^n$-type expressions: multiplying $n^n/n!$ (Problem 49, diverges to $\infty$) by $e^{-n}$ tames the growth all the way down to a merely-divergent $p=\frac12$ series, rather than to convergence.
Result
The series diverges: