Determine the limit to which the following expression
Determine the limit to which the following expression tends as $n \to \infty$: $$ a_n = \frac{1}{n}\sum_{k=1}^{n} \ln k. $$
Explanation of Method
Summing logarithms is the same as taking the log of a product, so $\sum_{k=1}^n \ln k = \ln(n!)$, turning the problem into finding $\lim \frac{\ln(n!)}{n}$. Stirling's approximation gives $\ln(n!)\sim n\ln n - n$, and dividing by $n$ shows the expression grows like $\ln n - 1$ — unboundedly, but only logarithmically slowly. A Riemann-sum argument gives an independent confirmation of the same result.
- Rewrite the sum as $\ln(n!)$ using logarithm properties.
- Apply Stirling's approximation and divide by $n$.
- Confirm the result via a Riemann-sum interpretation.
Step 1 · Recognize the sum as a logarithm of a product
$$ \sum_{k=1}^{n}\ln k = \ln\!\left(\prod_{k=1}^{n} k\right) = \ln(n!), \qquad\text{so}\qquad a_n = \frac{\ln(n!)}{n}. $$
Step 2 · Apply Stirling's approximation
Stirling's formula gives $\ln(n!) = n\ln n - n + \frac12\ln(2\pi n) + O\left(\frac1n\right)$. Dividing by $n$:
$$ \frac{\ln(n!)}{n} = \ln n - 1 + \frac{\ln(2\pi n)}{2n} + O\left(\frac1{n^2}\right). $$
Since $\frac{\ln(2\pi n)}{2n}\to0$ as $n\to\infty$:
$$ \lim_{n\to\infty} a_n = \lim_{n\to\infty}(\ln n - 1) = \infty. $$
Step 3 · Confirm via a Riemann sum
Writing $\ln k = \ln\!\left(\frac{k}{n}\cdot n\right) = \ln\!\left(\frac{k}{n}\right)+\ln n$:
$$ a_n = \frac1n\sum_{k=1}^{n}\ln\!\left(\frac kn\right) + \ln n. $$
The first term is a Riemann sum for $\int_0^1 \ln x\,dx = [x\ln x - x]_0^1 = -1$. So $a_n \to -1 + \ln n \to \infty$, confirming the Stirling-based result and giving the sharper asymptotic $a_n \sim \ln n - 1$.
Summary Table
| Quantity | Asymptotic behavior |
|---|---|
| $\ln(n!)$ | $\sim n\ln n - n$ |
| $a_n = \dfrac{\ln(n!)}{n}$ | $\sim \ln n - 1 \to \infty$ |
| Riemann-sum check | $\int_0^1 \ln x\,dx = -1$, consistent |
Additional Notes
- Although $a_n\to\infty$, the growth is purely logarithmic — extremely slow. Reaching $a_n>10$, for instance, requires $n$ on the order of $e^{11}\approx60{,}000$.
- This problem is the natural counterpart to Problem 16 ($\sqrt[n]{n!}\sim n/e$): exponentiating $a_n = \ln(n!)/n \sim \ln n - 1$ gives $(n!)^{1/n} = e^{a_n} \sim e^{\ln n - 1} = n/e$, the exact same asymptotic used there.
- The improper integral $\int_0^1 \ln x\,dx=-1$ converges despite $\ln x\to-\infty$ as $x\to0^+$, because $x\ln x\to0$ fast enough; this is the same phenomenon that keeps the Riemann-sum argument valid near $k=1$ (i.e. $x$ near $0$).
Result
The limit of the expression is: