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: $$ \sum_{n=1}^{\infty} \frac{n}{\sqrt[n]{n!}}. $$
Explanation of Method
Taking the $n$-th root of Stirling's formula shows $(n!)^{1/n}\sim\frac ne$, since the correction factor $(2\pi n)^{1/(2n)}\to1$. Substituting this into the general term gives $\frac n{(n!)^{1/n}}\sim\frac n{n/e}=e$ — a nonzero constant limit rather than $0$ — so the Divergence Test settles convergence immediately.
- Take the $n$-th root of Stirling's formula and simplify each factor.
- Show $(n!)^{1/n}\sim n/e$.
- Apply the Divergence Test to the resulting limit.
Step 1 · Take the $n$-th root of Stirling's formula
$$ (n!)^{1/n} = (2\pi n)^{1/(2n)}\cdot n\cdot e^{-1+\theta_n/(12n^2)}. $$
As $n\to\infty$: $\ln\left[(2\pi n)^{1/(2n)}\right]=\dfrac{\ln(2\pi n)}{2n}\to0$, so $(2\pi n)^{1/(2n)}\to1$; and since $0<\theta_n<1$, $\dfrac{\theta_n}{12n^2}\to0$, so $e^{-1+\theta_n/(12n^2)}\to e^{-1}$.
Step 2 · Asymptotic behavior of $(n!)^{1/n}$
Combining these:
$$ (n!)^{1/n} \sim 1\cdot n\cdot\frac1e = \frac ne. $$
Step 3 · Compute the general term and apply the divergence test
$$ a_n = \frac{n}{(n!)^{1/n}} \sim \frac{n}{n/e} = e. $$
More precisely, $\lim_{n\to\infty}\dfrac{n}{(n!)^{1/n}}=e\approx2.718\ne0$. By the Divergence Test, $\sum_{n=1}^{\infty}\dfrac n{\sqrt[n]{n!}}$ diverges.
Summary Table
| Quantity | Value |
|---|---|
| $(n!)^{1/n}$ | $\sim n/e$ |
| $a_n = n/(n!)^{1/n}$ | $\to e$ |
| Conclusion | Diverges (Divergence Test) |
Additional Notes
- This is the exact same asymptotic fact used in Problem 16, obtained there via direct Stirling-approximation manipulation and here via the explicit form with the $\theta_n$ error term — both confirm $(n!)^{1/n}\sim n/e$ with equal rigor.
- The equivalent logarithmic computation, $\dfrac{\ln(n!)}n = \ln n-1+\dfrac{\ln(2\pi n)}{2n}+\dfrac{\theta_n}{12n^2} \to \ln n-1$, gives $(n!)^{1/n}=e^{\ln(n!)/n}\sim e^{\ln n-1}=n/e$, an alternative derivation route matching Problem 19's approach to $\frac1n\sum\ln k$.
- Because the general term tends to the fixed constant $e$ rather than to $0$ or $\infty$, this series is a clean illustration of the Divergence Test in its simplest form: the partial sums grow essentially linearly, like $en$, once $n$ is large.
Result
The series diverges: