Nice Intermediate

The sum is the harmonic series, which can be approximated by: $\sum_{k=1}^{n} \frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}=\ln n+\gamma$ where $\gamma$ is the Euler–Mascheroni constant. $\gamma\approx0.577\dots$

Therefore, we can rewrite the question: $\begin{aligned}\lim_{n\to\infty}\frac{1}{\ln n}\sum_{k=1}^{n}\frac{1}{k}&=\lim_{n\to\infty}\frac{1}{\ln n}\left(\ln n+\gamma\right)\\ &=\lim_{n\to\infty}\left(1+\frac{\gamma}{\ln n}\right)\\ &=1\end{aligned}$ As $n$ approaches infinity, $\ln n$ also approaches infinity, so the $\frac{\gamma}{\ln n}$ approaches zero and disappears, leaving an answer of $\boxed{1}$