问题：求\lim\limits_{n\to\infty} n^2\sum\limits_{k=1}^{n}\dfrac{\mathrm{e}^{k/n}}{n^3+k^2} .
注意到\dfrac{\mathrm{e}^{k/n}}{n^3+n^2}\leqslant\dfrac{\mathrm{e}^{k/n}}{n^3+k^2}\leqslant\dfrac{\mathrm{e}^{k/n}}{n^3+n^2}.
故\dfrac{1}{n^3+n^2}\cdot\dfrac{\mathrm{e}^{1/n}(1-\mathrm{e})}{1-\mathrm{e}^{1/n}}=\sum\limits_{k=1}^{n}\dfrac{\mathrm{e}^{k/n}}{n^3+n^2}\leqslant\sum\limits_{k=1}^{n}\dfrac{\mathrm{e}^{k/n}}{n^3+k^2}\leqslant\sum\limits_{k=1}^{n}\dfrac{\mathrm{e}^{k/n}}{n^3+n^2}=\dfrac{1}{n^3}\cdot\dfrac{\mathrm{e}^{1/n}(1-\mathrm{e})}{1-\mathrm{e}^{1/n}}.
进而\dfrac{1}{n+1}\cdot\dfrac{\mathrm{e}^{1/n}(1-\mathrm{e})}{1-\mathrm{e}^{1/n}}\leqslant n^2\sum\limits_{k=1}^{n}\dfrac{\mathrm{e}^{k/n}}{n^3+k^2}\leqslant \dfrac{1}{n}\cdot\dfrac{\mathrm{e}^{1/n}(1-\mathrm{e})}{1-\mathrm{e}^{1/n}}.
而\lim\limits_{n\to\infty}\dfrac{1}{n}\cdot\dfrac{\mathrm{e}^{1/n}(1-\mathrm{e})}{1-\mathrm{e}^{1/n}} = \lim\limits_{n\to\infty}\dfrac{1}{n+1}\cdot\dfrac{\mathrm{e}^{1/n}(1-\mathrm{e})}{1-\mathrm{e}^{1/n}}=\mathrm{e}-1.
故原式=\mathrm{e}-1.
