Using Raabe's test, examine the convergence of: sum
Using Raabe's test, examine the convergence of: $$ \sum_{n=1}^{\infty} \frac{1}{n!}\left(\frac{n}{e}\right)^n. $$
Explanation of Method
D'Alembert's ratio test gives exactly $1$ here, since $u_n/u_{n+1}=e\left(\frac n{n+1}\right)^n\to e\cdot e^{-1}=1$; Raabe's test therefore needs the next-order correction. Using the refined asymptotic expansion $\left(1+\frac1n\right)^n=e\cdot e^{-1/(2n)+O(1/n^2)}$ isolates a correction term of order $\frac1{2n}$, giving Raabe limit $r=\frac12<1$ and hence divergence — consistent with Stirling's formula, which shows the terms behave like $\frac1{\sqrt{2\pi n}}$.
- Compute the plain ratio $u_n/u_{n+1}=e\left(\frac n{n+1}\right)^n$, which tends to $1$.
- Refine using the expansion $\left(1+\frac1n\right)^n=e\left(1-\frac1{2n}+O(1/n^2)\right)$.
- Evaluate the Raabe limit $r=\frac12<1$ and confirm via Stirling's formula.
Step 1 · Compute the plain ratio
With $u_n=\dfrac{n^n}{e^n\,n!}$:
$$ \frac{u_n}{u_{n+1}} = \frac{n^n\cdot e\cdot(n+1)}{(n+1)^{n+1}} = e\left(\frac{n}{n+1}\right)^n. $$
Step 2 · Refine the asymptotics
Using $\left(1+\frac1n\right)^n = e\cdot\exp\left(-\frac1{2n}+O(1/n^2)\right) = e\left(1-\frac1{2n}+O(1/n^2)\right)$:
$$ \left(\frac{n}{n+1}\right)^n = \frac1e\left(1+\frac1{2n}+O(1/n^2)\right) \implies \frac{u_n}{u_{n+1}} = 1+\frac1{2n}+O(1/n^2). $$
Step 3 · Apply Raabe's test
$$ r = \lim_{n\to\infty} n\left(\frac{u_n}{u_{n+1}}-1\right) = \lim_{n\to\infty}\left(\frac12+O(1/n)\right) = \frac12. $$
Since $r=\frac12<1$, by Raabe's test the series diverges. This matches Stirling's formula $n!\sim\sqrt{2\pi n}\left(\frac ne\right)^n$, giving $u_n\sim\dfrac1{\sqrt{2\pi n}}$, and $\sum\frac1{\sqrt n}$ diverges.
Summary Table
| Quantity | Value |
|---|---|
| $\lim u_n/u_{n+1}$ (D'Alembert) | $1$ (inconclusive) |
| Raabe limit $r$ | $1/2$ |
| Asymptotic $u_n$ (Stirling) | $\sim1/\sqrt{2\pi n}$ |
| Conclusion | Diverges |
Additional Notes
- This is a classic case where D'Alembert's ratio test is completely uninformative ($\to1$ exactly), yet the series' true behavior — divergence like $1/\sqrt n$ — is easily read off from Stirling's formula alone, without any Raabe computation at all; the Raabe analysis here mainly serves to derive that same fact from first principles.
- The correction coefficient $\frac12$ appearing in the Raabe limit is exactly half of the corresponding correction in $\left(1+\frac1n\right)^n=e-\frac e{2n}+O(1/n^2)$, tracing directly back to the second term of the standard asymptotic expansion for $e$.
- This series is essentially the reciprocal-Stirling normalization $n!\left(\frac en\right)^n$ turned upside down; its divergence at rate $n^{-1/2}$ is the same rate seen throughout the Stirling-based divergence problems in the 50s (e.g. Problems 51, 52), reflecting a shared underlying $\sqrt{2\pi n}$ factor.
Result
The series diverges: