方企勤 第四章 级 数 第8题
📝 题目
证(1)把 $\displaystyle{\left\lbrack {{a}_{1}, + \infty }\right\rbrack}$ 用分点 ${a}_{1} = {S}_{1} < {S}_{2} < \cdots < {S}_{k} < {S}_{k + 1} < \cdots$ 分成无限个小区间,在 $\left\lbrack {{S}_{k},{S}_{k + 1}}\right\rbrack$ 上,因 ${S}_{k + 1} - {S}_{k} = {a}_{k + 1}$ 及 $1/x$ 单调性, 我们有
$$ \frac{{a}_{k + 1}}{{S}_{k}} \geq {\int }_{{S}_{k}}^{{S}_{k + 1}}\frac{\mathrm{d}x}{x}\;\left( {k = 1,2,\cdots }\right) , $$
从而
$$ \mathop{\sum }\limits_{{k = 1}}^{n}\frac{{a}_{k + 1}}{{S}_{k}} \geq {\int }_{{a}_{1}}^{{S}_{n}}\frac{\mathrm{d}x}{x}\;\left( {\forall n \in N}\right) . $$
当 $\displaystyle{n \rightarrow \infty}$ 时, $\displaystyle{+ \infty \geq \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{a}_{n + 1}}{{S}_{n}} \geq {\int }_{{a}_{1}}^{\infty }\frac{\mathrm{d}x}{x} = + \infty}$ ,即得结论.
(2)我们考虑级数 $\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{a}_{n} - {a}_{n + 1}}{{S}_{n}}}$ ,因 $\frac{1}{{S}_{n}}$ 单调下降且趋于 0,及 $\left| {\mathop{\sum }\limits_{{k = 1}}^{\infty }\left( {{a}_{k} - {a}_{k + 1}}\right) }\right| \leq {2M}$ ,故级数 $\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{a}_{n} - {a}_{n + 1}}{{S}_{n}}}$ 收敛,于是由第 (1) 小题推出级数 $\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}/{S}_{n}}$ 发散.
又证 因对任意固定的 $\displaystyle{n,\mathop{\lim }\limits_{{p \rightarrow + \infty }}{S}_{n}/{S}_{n + p} = 0}$ ,所以 $\exists p > 0$ ,使 ${S}_{n}/{S}_{n + p} < 1/2$ . 于是对 ${\varepsilon }_{0} = 1/2 > 0,\forall N \in N,\exists n = N + 1 > N$ ,有
$$ \frac{{a}_{n + 1}}{{S}_{n + 1}} + \frac{{a}_{n + 2}}{{S}_{n + 2}} + \cdots + \frac{{a}_{n + p}}{{S}_{n + p}} > \frac{{S}_{n + p} - {S}_{n}}{{S}_{n + p}} = 1 - \frac{{S}_{n}}{{S}_{n + p}} > \frac{1}{2}. $$
故由收敛原理知 $\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}/{S}_{n}}$ 发散.
💡 答案解析
证(1)把 $\displaystyle{\left\lbrack {{a}_{1}, + \infty }\right\rbrack}$ 用分点 ${a}_{1} = {S}_{1} < {S}_{2} < \cdots < {S}_{k} < {S}_{k + 1} < \cdots$ 分成无限个小区间,在 $\left\lbrack {{S}_{k},{S}_{k + 1}}\right\rbrack$ 上,因 ${S}_{k + 1} - {S}_{k} = {a}_{k + 1}$ 及 $1/x$ 单调性, 我们有
$$ \frac{{a}_{k + 1}}{{S}_{k}} \geq {\int }_{{S}_{k}}^{{S}_{k + 1}}\frac{\mathrm{d}x}{x}\;\left( {k = 1,2,\cdots }\right) , $$
从而
$$ \mathop{\sum }\limits_{{k = 1}}^{n}\frac{{a}_{k + 1}}{{S}_{k}} \geq {\int }_{{a}_{1}}^{{S}_{n}}\frac{\mathrm{d}x}{x}\;\left( {\forall n \in N}\right) . $$
当 $\displaystyle{n \rightarrow \infty}$ 时, $\displaystyle{+ \infty \geq \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{a}_{n + 1}}{{S}_{n}} \geq {\int }_{{a}_{1}}^{\infty }\frac{\mathrm{d}x}{x} = + \infty}$ ,即得结论.
(2)我们考虑级数 $\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{a}_{n} - {a}_{n + 1}}{{S}_{n}}}$ ,因 $\frac{1}{{S}_{n}}$ 单调下降且趋于 0,及 $\left| {\mathop{\sum }\limits_{{k = 1}}^{\infty }\left( {{a}_{k} - {a}_{k + 1}}\right) }\right| \leq {2M}$ ,故级数 $\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{a}_{n} - {a}_{n + 1}}{{S}_{n}}}$ 收敛,于是由第 (1) 小题推出级数 $\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}/{S}_{n}}$ 发散.
又证 因对任意固定的 $\displaystyle{n,\mathop{\lim }\limits_{{p \rightarrow + \infty }}{S}_{n}/{S}_{n + p} = 0}$ ,所以 $\exists p > 0$ ,使 ${S}_{n}/{S}_{n + p} < 1/2$ . 于是对 ${\varepsilon }_{0} = 1/2 > 0,\forall N \in N,\exists n = N + 1 > N$ ,有
$$ \frac{{a}_{n + 1}}{{S}_{n + 1}} + \frac{{a}_{n + 2}}{{S}_{n + 2}} + \cdots + \frac{{a}_{n + p}}{{S}_{n + p}} > \frac{{S}_{n + p} - {S}_{n}}{{S}_{n + p}} = 1 - \frac{{S}_{n}}{{S}_{n + p}} > \frac{1}{2}. $$
故由收敛原理知 $\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}/{S}_{n}}$ 发散.