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高中数学选择性必修 第二册（381题）
设数列
令
知识点：第四章 数列
参考答案：证明： \({b_n} = {a_n}^2 = {\left( {2n + 1} \right)^2} = 4{n^2} + 4n + 1 > 4n\left( {n + 1} \right)\) ， \(\therefore \frac{1}{{{b_n}}} < \frac{1}{{4n(n + 1)}} = \frac{1}{4}(\frac{1}{n} - \frac{1}{{n + 1}})\) 则 \(\frac{1}{{{b_1}}} + \frac{1}{{{b_2}}} + \frac{1}{{{b_3}}} + \cdot \cdot \cdot + \frac{1}{{{b_n}}} < \frac{1}{4}\left[ {\left( {1 - \frac{1}{2}} \right) + \left( {\frac{1}{2} - \frac{1}{3}} \right) + \cdot \cdot \cdot + \left( {\frac{1}{n} - \frac{1}{{n + 1}}} \right)} \right] = \frac{1}{4}\left( {1 - \frac{1}{{n + 1}}} \right) < \frac{1}{4}\) ．