how to show convergence and evaluate $n\sum_{k=1}^{2n} \frac{e^{-n/k}}{k^2}$

Question

I tried this : $$\lim\limits_{n \rightarrow +\infty}n\sum_{k=1}^{2n} \frac{e^{-n/k}}{k^2} = \int_{0}^{1} \frac{e^{-1/x}}{x^2} dx$$ to evaluate, and to show convergence, I think $$n\sum_{k=1}^{2n} \frac{e^{-n/k}}{k^2} < \sum_{k>1}{} \frac{1}{k^2}$$ which converges.

Answer 1

$$ n\sum\limits_{k = 1}^{2n} {\frac{{e^{ - \frac{n}{k}} }}{{k^2 }}} = \frac{1}{2}\frac{1}{{2n}}\sum\limits_{k = 1}^{2n} {\frac{{e^{ - \frac{1}{2}\frac{{2n}}{k}} }}{{(k/2n)^2 }}} \to \frac{1}{2}\int_0^1 {\frac{{e^{ - 1/(2x)} }}{{x^2 }}dx} \mathop = \limits^{t = 1/(2x)} \int_{1/2}^{ + \infty } {e^{ - t} dt} = e^{ - 1/2} . $$