📝 题目
例 1 求证: $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}\sqrt[n]{n} = 1}$ .
💡 答案与解析
证 因为
$$ 1 \leq \sqrt[n]{n} = {\left( \sqrt{n} \cdot \sqrt{n} \cdot \overset{n - 2}{\overbrace{1 \cdot \cdots \cdot 1}}\right) }^{\frac{1}{n}} $$
$$ \leq \frac{\sqrt[n]{n} + \sqrt{n + 1 + \cdots + 1}}{n} $$
$$ = \frac{2\sqrt{n} + n - 2}{n} < 1 + \frac{2}{\sqrt{n}}, $$
所以
$$ \left| {\sqrt[n]{n} - 1}\right| < \frac{2}{\sqrt{n}} $$
于是,对任给定 $\varepsilon > 0$ ,取 $N = \left\lbrack \frac{4}{{\varepsilon }^{2}}\right\rbrack + 1$ ,当 $n > N$ 时便有
$$ \left| {\sqrt[n]{n} - 1}\right| < \frac{2}{\sqrt{n}} < \frac{2}{\sqrt{N}} < \varepsilon . $$