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【微信公众号：墨尘的数学笔记】
例4：已知数列 \{a_n\} 满足 a_1 = 2 , a_{n+1} = 2\left(1 + \frac{1}{n}\right)^2 a_n , n \in N_+
(1) 求证：数列 \left\{\frac{a_n}{n^2}\right\} 是等比数列，并求出数列 \{a_n\} 的通项公式
(2) 设 c_n = \frac{n}{a_n} ，求证： c_1 + c_2 + \cdots + c_n < \frac{17}{24}
\begin{aligned}\text{解：(1)} \quad a_{n+1} &= 2\left(1 + \frac{1}{n}\right)^2 a_n = 2 \cdot \frac{(n+1)^2}{n^2} a_n \\ \therefore \frac{a_{n+1}}{(n+1)^2} &= 2 \cdot \frac{a_n}{n^2} \quad \therefore \left\{\frac{a_n}{n^2}\right\} \text{是公比为2的等比数列} \\ \therefore \frac{a_n}{n^2} &= \left(\frac{a_1}{1^2}\right) \cdot 2^{n-1} = 2^n \\ \therefore a_n &= n^2 \cdot 2^n\end{aligned}
(2) 思路： c_n = \frac{n}{a_n} = \frac{1}{n \cdot 2^n} ，无法直接求和，所以考虑放缩成为可求和的通项公式（不等号： < ）。若要放缩为裂项相消的形式，那么需要构造出“顺序同构”的特点。观察分母中有 n ，故分子分母通乘以 (n-1) ，再进行放缩调整为裂项相消形式。
\begin{aligned}\text{解：} c_n &= \frac{n}{a_n} = \frac{1}{n \cdot 2^n} = \frac{n-1}{n(n-1)2^n} \\ \text{而} \quad \frac{1}{(n-1)2^{n-1}} - \frac{1}{n \cdot 2^n} &= \frac{2n - (n-1)}{n(n-1)2^n} = \frac{n+1}{n(n-1)2^n} \\ \text{所以} \quad c_n &= \frac{n-1}{n(n-1)2^n} < \frac{n+1}{n(n-1)2^n} = \frac{1}{(n-1)2^{n-1}} - \frac{1}{n \cdot 2^n} \quad (n \ge 2)\end{aligned}
\begin{aligned}c_1 + c_2 + \cdots + c_n &< c_1 + c_2 + c_3 + \left( \frac{1}{3 \cdot 2^3} - \frac{1}{4 \cdot 2^4} + \frac{1}{4 \cdot 2^4} - \frac{1}{5 \cdot 2^5} + \cdots + \frac{1}{(n-1)2^{n-1}} - \frac{1}{n \cdot 2^n} \right) \\ &= \frac{1}{2} + \frac{1}{8} + \frac{1}{24} + \frac{1}{24} - \frac{1}{n \cdot 2^n} = \frac{17}{24} - \frac{1}{n \cdot 2^n} < \frac{17}{24} \quad (n > 3) \\ \therefore c_n &> 0 \quad \therefore c_1 < c_1 + c_2 < c_1 + c_2 + c_3 = \frac{16}{24} < \frac{17}{24}\end{aligned}
小炼有话说：(1) 本题先确定放缩的类型，向裂项相消放缩，从而按“依序同构”的目标进
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