a) Ta có: \(\displaystyle {S_1} = \frac{1}{n}.{e^{ - \frac{1}{n}}}\); \(\displaystyle {S_2} = \frac{1}{n}.{e^{ - \frac{2}{n}}}\); …;\(\displaystyle {S_n} = \frac{1}{n}.{e^{ - \frac{n}{n}}}\)
\(\displaystyle \Rightarrow {S_n} = \frac{1}{n}\left( {{e^{ - \frac{1}{n}}} + {e^{ - \frac{2}{n}}} + ... + {e^{ - \frac{n}{n}}}} \right)\)\(\displaystyle = \frac{1}{n}.{e^{ - \frac{1}{n}}}\frac{{1 - {{\left( {{e^{ - \frac{1}{n}}}} \right)}^n}}}{{1 - {e^{ - \frac{1}{n}}}}} = \frac{1}{n}.\frac{{1 - {e^{ - 1}}}}{{{e^{\frac{1}{n}}} - 1}}\)
b) \(\displaystyle \mathop {\lim }\limits_{n \to \infty } {S_n} = 1 - {e^{ - 1}}\)
Mặt khác \(\displaystyle S = \int\limits_0^1 {{e^{ - x}}dx} = - \left. {{e^{ - x}}} \right|_0^1 = 1 - {e^{ - 1}}\).
Do đó \(\displaystyle \mathop {\lim }\limits_{n \to \infty } {S_n} = 1 - {e^{ - 1}} = \int\limits_0^1 {{e^{ - x}}dx} = S\)
