Bài tập 3.35 trang 178 SBT Toán 12

a) Ta có: 

\(\begin{array}{l}
{S_1} = \frac{1}{n}.{e^{ - \frac{1}{n}}}\\
{S_2} = \frac{1}{n}.{e^{ - \frac{2}{n}}}\\
,...,{S_n} = \frac{1}{n}.{e^{ - \frac{n}{n}}}
\end{array}\)

\(\begin{array}{l}
 \Rightarrow {S_n} = \frac{1}{n}\left( {{e^{ - \frac{1}{n}}} + {e^{ - \frac{2}{n}}} + ... + {e^{ - \frac{n}{n}}}} \right)\\
 = \frac{1}{n}.{e^{ - \frac{1}{n}}}\frac{{1 - {{\left( {{e^{ - \frac{1}{n}}}} \right)}^n}}}{{1 - {e^{ - \frac{1}{n}}}}} = \frac{1}{n}.\frac{{1 - {e^{ - 1}}}}{{{e^{\frac{1}{n}}} - 1}}
\end{array}\)

b) \(\mathop {\lim }\limits_{n \to \infty } {S_n} = 1 - {e^{ - 1}}\)

Mặt khác: \(S = 1\int\limits_0^1 {{e^{ - x}}} dx = \left. { - {e^{ - x}}} \right|_0^1 = 1 - {e^{ - 1}}\)

Do đó \(\mathop {\lim }\limits_{n \to \infty } {S_n} = 1 - {e^{ - 1}} = \int \limits_0^1 {e^{ - x}}dx = S\)

-- Mod Toán 12 HỌC247