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: $$ \sum_{n=1}^{\infty} \frac{\ln n!}{n \ln n}. $$
Explanation of Method
Taking the logarithm of Stirling's formula gives $\ln n! = n\ln n - n + \frac12\ln n + O(1)$, so the dominant term is exactly $n\ln n$ — the same as the denominator. Dividing through shows the general term tends to $1$, not $0$, so the Divergence Test settles the question immediately.
- Take logs of Stirling's formula to expand $\ln n!$.
- Divide by $n\ln n$ and simplify each piece.
- Apply the Divergence Test to the resulting limit of $1$.
Step 1 · Expand $\ln n!$ via Stirling's formula
$$ \ln(n!) = \frac12\ln(2\pi n) + n\ln n - n + \frac{\theta_n}{12n} = n\ln n - n + \frac12\ln n + O(1). $$
Step 2 · Divide by $n\ln n$
$$ a_n = \frac{\ln n!}{n\ln n} = \frac{n\ln n-n+\frac12\ln n+O(1)}{n\ln n} = 1 - \frac1{\ln n} + \frac1{2n} + O\!\left(\frac1{n\ln n}\right). $$
Step 3 · Apply the divergence test
As $n\to\infty$: $\frac1{\ln n}\to0$, $\frac1{2n}\to0$, so:
$$ \lim_{n\to\infty} a_n = 1-0+0 = 1 \neq 0. $$
By the Divergence Test, $\sum_{n=1}^{\infty}\dfrac{\ln n!}{n\ln n}$ diverges.
Summary Table
| Quantity | Value |
|---|---|
| $\ln n!$ asymptotics | $n\ln n - n + \frac12\ln n + O(1)$ |
| General term $a_n$ | $1-\dfrac1{\ln n}+O\!\left(\dfrac1n\right)$ |
| $\lim_{n\to\infty}a_n$ | $1$ |
| Conclusion | Diverges |
Additional Notes
- Numerically the terms approach $1$ quite slowly, since the correction is $-\frac1{\ln n}$: even at $n=1000$, $a_n\approx1-\frac1{6.9}\approx0.86$, still visibly below $1$.
- This problem is the natural companion to Problem 19 ($\frac1n\sum\ln k=\frac{\ln n!}n\sim\ln n-1\to\infty$): dividing that unbounded quantity by an extra factor of $\ln n$ here tames the growth down to a finite (but nonzero) limit.
- Since the terms converge to the nonzero constant $1$ rather than merely failing to shrink fast enough, this is a case where the Divergence Test applies directly, without needing any comparison series.
Result
The series diverges: