📝 题目
1.3.8 求证: 如下序列的极限存在:
$$ \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {1 + \frac{1}{{2}^{2}}}\right) \left( {1 + \frac{1}{{3}^{2}}}\right) \cdots \left( {1 + \frac{1}{{n}^{2}}}\right) . $$
💡 答案与解析
1.3.8 ${\left( 1 + \frac{1}{{k}^{2}}\right) }^{{k}^{2}} < \mathrm{e}\overset{\text{ 两边取对数 }}{ = }{k}^{2}\ln \left( {1 + \frac{1}{{k}^{2}}}\right) < 1 \Rightarrow 1 + \frac{1}{{k}^{2}} < {\mathrm{e}}^{\frac{1}{{k}^{2}}}$
$$ \Rightarrow \mathop{\prod }\limits_{{k = 2}}^{n}\left( {1 + \frac{1}{{k}^{2}}}\right) < {\mathrm{e}}^{\mathop{\sum }\limits_{{k = 2}}^{n}\frac{1}{{k}^{2}}}. $$
再注意到
$$ \frac{1}{{k}^{2}} < \frac{1}{k\left( {k - 1}\right) }\left( {k \geq 2}\right) \Rightarrow \mathop{\sum }\limits_{{k = 2}}^{n}\frac{1}{{k}^{2}} < \mathop{\sum }\limits_{{k = 2}}^{n}\left( {\frac{1}{k - 1} - \frac{1}{k}}\right) $$
$$ = 1 - \frac{1}{n} < 1\left( {n > 2}\right) , $$
故有
$$ \mathop{\prod }\limits_{{k = 2}}^{n}\left( {1 + \frac{1}{{k}^{2}}}\right) < \mathrm{e} $$