📝 题目
证明 我们有
$$ \left| {{x}_{n + p} - {x}_{n}}\right| = \frac{1}{{\left( n + 1\right) }^{2}} + \cdots + \frac{1}{{\left( n + p\right) }^{2}} $$
$$ < \frac{1}{n\left( {n + 1}\right) } + \cdots + \frac{1}{\left( {n + p - 1}\right) \left( {n + p}\right) } $$
$$ = \left( {\frac{1}{n} - \frac{1}{n + 1}}\right) + \cdots + \left( {\frac{1}{n + p - 1} - \frac{1}{n + p}}\right) $$
$$ = \frac{1}{n} - \frac{1}{n + p} < \frac{1}{n}. $$
对任意 $\varepsilon > 0$ ,可取 $N = \left\lbrack {1/\varepsilon }\right\rbrack + 1$ ,则对任意的 $n > N$ 和 $p \in \mathbb{N}$ 都有
$$ \left| {{x}_{n + p} - {x}_{n}}\right| < 1/n < \varepsilon . $$
💡 答案与解析
证明 我们有
$$ \left| {{x}_{n + p} - {x}_{n}}\right| = \frac{1}{{\left( n + 1\right) }^{2}} + \cdots + \frac{1}{{\left( n + p\right) }^{2}} $$
$$ < \frac{1}{n\left( {n + 1}\right) } + \cdots + \frac{1}{\left( {n + p - 1}\right) \left( {n + p}\right) } $$
$$ = \left( {\frac{1}{n} - \frac{1}{n + 1}}\right) + \cdots + \left( {\frac{1}{n + p - 1} - \frac{1}{n + p}}\right) $$
$$ = \frac{1}{n} - \frac{1}{n + p} < \frac{1}{n}. $$
对任意 $\varepsilon > 0$ ,可取 $N = \left\lbrack {1/\varepsilon }\right\rbrack + 1$ ,则对任意的 $n > N$ 和 $p \in \mathbb{N}$ 都有
$$ \left| {{x}_{n + p} - {x}_{n}}\right| < 1/n < \varepsilon . $$