Based on the general term a_n, conclude that
Based on the general term $a_n$, conclude that the following series diverges: $$ \sum_{n=1}^{\infty} \frac{n}{\sqrt[n]{n!}}. $$
Explanation of Method
Evaluating $\lim_{n\to\infty} \frac{n}{(n!)^{1/n}}$ directly is not obvious, since both numerator and the $n$-th root of $n!$ grow with $n$. Stirling's approximation $n!\sim\sqrt{2\pi n}\,(n/e)^n$ resolves this: taking the $n$-th root shows $(n!)^{1/n}\sim n/e$ (the $\sqrt{2\pi n}$ factor contributes nothing in the limit, since its $n$-th root tends to $1$), which makes the ratio $n/(n!)^{1/n}$ tend to the constant $e$ rather than to zero.
- Apply Stirling's approximation to $(n!)^{1/n}$.
- Show the extra factor $(2\pi n)^{1/(2n)}$ tends to $1$.
- Compute the resulting limit of $a_n$ and apply the Divergence Test.
Step 1 · Identify the general term and apply Stirling's approximation
$$ a_n = \frac{n}{\sqrt[n]{n!}} = \frac{n}{(n!)^{1/n}}. $$
Stirling's approximation states $n! \sim \sqrt{2\pi n}\left(\dfrac{n}{e}\right)^n$ as $n\to\infty$. Taking the $n$-th root:
$$ (n!)^{1/n} \sim (2\pi n)^{1/(2n)} \cdot \frac{n}{e}. $$
Step 2 · Show $(2\pi n)^{1/(2n)} \to 1$
Taking logarithms, $\ln\left[(2\pi n)^{1/(2n)}\right] = \dfrac{\ln(2\pi n)}{2n} \to 0$ as $n\to\infty$, since logarithmic growth is far slower than linear. Hence $(2\pi n)^{1/(2n)} \to e^0 = 1$, and so:
$$ (n!)^{1/n} \sim \frac{n}{e}. $$
Step 3 · Compute the limit of $a_n$ and apply the divergence test
$$ \lim_{n\to\infty} a_n = \lim_{n\to\infty} \frac{n}{(n!)^{1/n}} = \lim_{n\to\infty} \frac{n}{n/e} = e. $$
Since $\lim_{n\to\infty} a_n = e \approx 2.718 \neq 0$, by the Divergence Test the series $\sum_{n=1}^{\infty} \frac{n}{\sqrt[n]{n!}}$ diverges.
Summary Table
| Quantity | Asymptotic behavior |
|---|---|
| $(n!)^{1/n}$ | $\sim \dfrac{n}{e}$ |
| $a_n = \dfrac{n}{(n!)^{1/n}}$ | $\to e$ |
| Conclusion | Diverges (Divergence Test) |
Additional Notes
- The constant $e$ appearing here is not a coincidence — it is the same $e$ that shows up throughout factorial asymptotics, ultimately traceable to the ratio test limit $\lim (n+1)!/n!^{(n+1)/n} \cdot(\text{stuff})$ underlying Stirling's formula.
- A quicker route to the same conclusion uses the ratio-based limit $\lim_{n\to\infty}(n!)^{1/n}/n = 1/e$, a standard fact equivalent to Stirling's approximation, avoiding the need to track the $\sqrt{2\pi n}$ correction term explicitly.
- This problem is a good illustration that "$n$-th root of $n!$" grows linearly (like $n/e$) rather than like $n!$ itself, which is why the ratio $n/(n!)^{1/n}$ settles to a finite nonzero constant instead of vanishing or blowing up.
Result
The series diverges because: