📝 题目
例 2 求证: $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{1}{\sqrt[n]{n!}} = 0}$ .
💡 答案与解析
证 考虑二次函数 $f\left( x\right) = x\left( {n + 1 - x}\right) \left( {1 \leq x \leq n}\right)$ ,如图 1.3 所示. 显然,当 $1 \leq x \leq n$ 时, $x\left( {n + 1 - x}\right) \geq n$ . 故有
$$ {\left( n!\right) }^{2} = \left( {1 \cdot n}\right) \left( {2 \cdot \left( {n - 1}\right) }\right) \left( {3 \cdot \left( {n - 2}\right) }\right) \cdots \left( {n \cdot 1}\right) $$
$$ \geq \underset{n\text{ 个 }}{\underbrace{n \cdot n \cdot n \cdot \cdots \cdot n}} = {n}^{n} $$
$$ \Rightarrow \sqrt[n]{n!} \geq \sqrt{n} \Rightarrow \frac{1}{\sqrt[n]{n!}} \leq \frac{1}{\sqrt{n}}. $$
于是,对任给定 $\varepsilon > 0$ ,取 $N = \left\lbrack \frac{1}{{\varepsilon }^{2}}\right\rbrack + 1$ ,当 $n > N$ 时,便有
$$ 0 < \frac{1}{\sqrt[n]{n!}} < \frac{1}{\sqrt{n}} < \frac{1}{\sqrt{N}} < \varepsilon $$
\begin{center} \includegraphics[max width=0.2\textwidth]{images/003.jpg} \end{center} \hspace*{3em}
图 1.3