Limit Of Sum Divided By Natural Logarithm

Relevant wiki: Harmonic Number

$\large \begin{aligned} L & = \lim_{m \to \infty} \frac 1{\ln m}\sum_{n=2}^m \frac 1{n^{1+\frac 1{\ln n}}} \\ & = \lim_{m \to \infty} \frac 1{\ln m}\sum_{n=2}^m \frac 1{e^{\ln n+1}} \\ & = \lim_{m \to \infty} \frac 1{\ln m}\sum_{n=2}^m \frac 1{en} \\ & = \lim_{m \to \infty} \frac {{\color{#3D99F6}H_m} - 1}{e\ln m} & \small \color{#3D99F6} \text{where }H_m = \sum_{n=1}^n \frac 1n \text{, the }n\text{th harmonic number.} \\ & = \lim_{m \to \infty} \frac {{\color{#3D99F6}\ln m + \gamma} - 1}{e\ln m} & \small \color{#3D99F6} \text{where }\gamma \text{ is the Euler-Mascheroni constant.} \\ & = \boxed{\dfrac 1e} \end{aligned}$