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芝士点

夹逼定理:设\(\displaystyle\lim_{n \to \infty} a_n=\lim_{ \to \infty} b_n=a\),若\(\exists M\in N,当n>M时,恒有a_n\le c_n\le b_n,则 \displaystyle\lim_{n \to \infty}c_n=a\)

说人话:三胞胎,老大是1.1号出生,老三是1.3号出生,那么老二的出生时间一定在1.1号和1.3号之间。(老大和老三的出生时间如果无限接近,则老二出生时间可定)

你已经掌握了夹逼定理,试试看!


例题

求极限\(\displaystyle\lim_{n \to \infty} \frac{\displaystyle\sum_{k=1}^{n}\frac{1}{k}}{\ln n}\)

解 由于 \(y=\frac{1}{x}\)\(x>0\) 时单调减少, 则

\[ \begin{gathered} \int_k^{k+1} \frac{1}{x} \mathrm{~d} x \leqslant \frac{1}{k} \leqslant \int_{k-1}^k \frac{1}{x} \mathrm{~d} x \\ \int_1^{n+1} \frac{1}{x} \mathrm{~d} x=\sum_{k=1}^n \int_k^{k+1} \frac{1}{x} \mathrm{~d} x \leqslant \sum_{k=1}^n \frac{1}{k} \leqslant \sum_{k=2}^n \int_{k-1}^k \frac{1}{x} \mathrm{~d} x+1=\int_1^n \frac{1}{x} \mathrm{~d} x+1 \\ \text { 即 } \ln (n+1) \leqslant \sum_{k=1}^n \frac{1}{k} \leqslant \ln n+1 \Rightarrow \frac{\ln (n+1)}{\ln n} \leqslant \frac{\sum_{k=1}^n \frac{1}{k}}{\ln n} \leqslant \frac{\ln n+1}{\ln n} \\ \text { 而 } \lim _{n \rightarrow \infty} \frac{\ln (n+1)}{\ln n}=1=\lim _{n \rightarrow \infty} \frac{\ln n+1}{\ln n}, \text { 所以 } \lim _{n \rightarrow \infty} \frac{\sum_{k=1}^n \frac{1}{k}}{\ln n}=1 . \end{gathered} \]

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