《2025年高考总复习优化设计高中数学北师大版》

例3 (2022·新高考Ⅰ，17)记$S_n$为数列$\{ a_n\}$的前$n$项和，已知$a_1 = 1,\{\frac{S_n}{a_n}\}$是公差为$\frac{1}{3}$的等差数列.
(1)求$\{ a_n\}$的通项公式；
(2)证明：$\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}＜2$.
[课堂笔记]例3 (1)解（方法一）$\because \{\frac{S_n}{a_n}\}$是以$\frac{S_1}{a_1} = 1$为首项，以$\frac{1}{3}$为公差的等差数列，
$\therefore \frac{S_n}{a_n} = 1 + (n - 1) × \frac{1}{3} = \frac{n + 2}{3}$.
$\therefore S_n = \frac{n + 2}{3}a_n$.①
当$n \geq 2$时，$S_{n - 1} = \frac{n + 1}{3}a_{n - 1}$.②
① - ②得$a_n = \frac{n + 2}{3}a_n - \frac{n + 1}{3}a_{n - 1}$，
$\therefore \frac{n + 1}{3}a_{n - 1} = \frac{n - 1}{3}a_n$，$\therefore \frac{a_n}{a_{n - 1}} = \frac{n + 1}{n - 1}$，
$\therefore a_n = \frac{a_n}{a_{n - 1}} \cdot \frac{a_{n - 1}}{a_{n - 2}} \cdots \frac{a_2}{a_1} \cdot a_1 = \frac{n + 1}{n - 1} × \frac{n}{n - 2} × \frac{n - 1}{n - 3} × \cdots × \frac{4}{2} × \frac{3}{1} × a_1 (n \geq 2)$.
又$a_1 = 1$，$\therefore a_n = \frac{n(n + 1)}{2 × 1} × 1 = \frac{n(n + 1)}{2} (n \geq 2)$.
又当$n = 1$时，$a_1 = 1$也符合上式，
$\therefore a_n = \frac{n(n + 1)}{2}$.
（方法二）$\because \{\frac{S_n}{a_n}\}$是以$\frac{S_1}{a_1} = 1$为首项，以$\frac{1}{3}$为公差的等差数列，$\therefore \frac{S_n}{a_n} = 1 + (n - 1) × \frac{1}{3} = \frac{n + 2}{3} \cdot S_n = \frac{n + 2}{3}a_n$.①
当$n \geq 2$时，$S_{n - 1} = \frac{n + 1}{3}a_{n - 1}$.②
① - ②得$a_n = \frac{n + 2}{3}a_n - \frac{n + 1}{3}a_{n - 1}$，
$\therefore \frac{a_n}{n + 1} = \frac{a_{n - 1}}{n - 1} \cdots \frac{a_n}{a_{n - 1}} = \frac{n + 1}{n - 1} \cdot n$，
设$\frac{a_n}{n(n + 1)} = b_n$，则$b_n = b_{n - 1}$，
$\therefore \{b_n\}$为常数列，且$b_1 = \frac{a_1}{1 × 2} = \frac{1}{2}$，
$\therefore \frac{a_n}{n(n + 1)} = \frac{1}{2}$，$\therefore a_n = \frac{n(n + 1)}{2}$.
(2)证明 由(1)知，$\frac{1}{a_n} = \frac{2}{n(n + 1)} = 2(\frac{1}{n} - \frac{1}{n + 1})$，
$\therefore \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n} = 2(1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \cdots + \frac{1}{n} - \frac{1}{n + 1}) = 2(1 - \frac{1}{n + 1}) ＜ 2$.变式探究 证明 因为$\frac{1}{a_n} = \frac{2}{n(n + 2)} = \frac{1}{n} - \frac{1}{n + 2}$，
所以$\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n} = (1 - \frac{1}{3}) + (\frac{1}{2} - \frac{1}{4}) + (\frac{1}{3} - \frac{1}{5}) + \cdots + (\frac{1}{n - 1} - \frac{1}{n + 1}) + (\frac{1}{n} - \frac{1}{n + 2}) = 1 + \frac{1}{2} - \frac{1}{n + 1} - \frac{1}{n + 2} ＜ \frac{3}{2} ＜ \frac{n + 1}{n + 2} ＜ \frac{3}{2}$.
|变式探究|
(变条件变结论)若本例中$\{ a_n\}$的通项公式$a_n=\frac{n(n + 2)}{2}$，证明：$\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}＜\frac{3}{2}$.

答案： 例3
(1)解（方法一）$\because \{\frac{S_n}{a_n}\}$是以$\frac{S_1}{a_1} = 1$为首项，以$\frac{1}{3}$为公差的等差数列，
$\therefore \frac{S_n}{a_n} = 1 + (n - 1) × \frac{1}{3} = \frac{n + 2}{3}$.
$\therefore S_n = \frac{n + 2}{3}a_n$.①
当$n \geq 2$时，$S_{n - 1} = \frac{n + 1}{3}a_{n - 1}$.②
① - ②得$a_n = \frac{n + 2}{3}a_n - \frac{n + 1}{3}a_{n - 1}$，
$\therefore \frac{n + 1}{3}a_{n - 1} = \frac{n - 1}{3}a_n$，$\therefore \frac{a_n}{a_{n - 1}} = \frac{n + 1}{n - 1}$，
$\therefore a_n = \frac{a_n}{a_{n - 1}} \cdot \frac{a_{n - 1}}{a_{n - 2}} \cdots \frac{a_2}{a_1} \cdot a_1 = \frac{n + 1}{n - 1} × \frac{n}{n - 2} × \frac{n - 1}{n - 3} × \cdots × \frac{4}{2} × \frac{3}{1} × a_1 (n \geq 2)$.
又$a_1 = 1$，$\therefore a_n = \frac{n(n + 1)}{2 × 1} × 1 = \frac{n(n + 1)}{2} (n \geq 2)$.
又当$n = 1$时，$a_1 = 1$也符合上式，
$\therefore a_n = \frac{n(n + 1)}{2}$.
（方法二）$\because \{\frac{S_n}{a_n}\}$是以$\frac{S_1}{a_1} = 1$为首项，以$\frac{1}{3}$为公差的等差数列，$\therefore \frac{S_n}{a_n} = 1 + (n - 1) × \frac{1}{3} = \frac{n + 2}{3} \cdot S_n = \frac{n + 2}{3}a_n$.①
当$n \geq 2$时，$S_{n - 1} = \frac{n + 1}{3}a_{n - 1}$.②
① - ②得$a_n = \frac{n + 2}{3}a_n - \frac{n + 1}{3}a_{n - 1}$，
$\therefore \frac{a_n}{n + 1} = \frac{a_{n - 1}}{n - 1} \cdots \frac{a_n}{a_{n - 1}} = \frac{n + 1}{n - 1} \cdot n$，
设$\frac{a_n}{n(n + 1)} = b_n$，则$b_n = b_{n - 1}$，
$\therefore \{b_n\}$为常数列，且$b_1 = \frac{a_1}{1 × 2} = \frac{1}{2}$，
$\therefore \frac{a_n}{n(n + 1)} = \frac{1}{2}$，$\therefore a_n = \frac{n(n + 1)}{2}$.
(2)证明 由
(1)知，$\frac{1}{a_n} = \frac{2}{n(n + 1)} = 2(\frac{1}{n} - \frac{1}{n + 1})$，
$\therefore \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n} = 2(1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \cdots + \frac{1}{n} - \frac{1}{n + 1}) = 2(1 - \frac{1}{n + 1}) ＜ 2$.
变式探究 证明 因为$\frac{1}{a_n} = \frac{2}{n(n + 2)} = \frac{1}{n} - \frac{1}{n + 2}$，
所以$\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n} = (1 - \frac{1}{3}) + (\frac{1}{2} - \frac{1}{4}) + (\frac{1}{3} - \frac{1}{5}) + \cdots + (\frac{1}{n - 1} - \frac{1}{n + 1}) + (\frac{1}{n} - \frac{1}{n + 2}) = 1 + \frac{1}{2} - \frac{1}{n + 1} - \frac{1}{n + 2} ＜ \frac{3}{2} ＜ \frac{n + 1}{n + 2} ＜ \frac{3}{2}$.