第七章 典型综合题分析 · 第6题

证法 1 记 $a = \frac{\sqrt{5} - 1}{2},b = \frac{\sqrt{5} + 1}{2}$ . 则

$$ f\left( x\right) = \frac{1}{\sqrt{5}}\left\{ {\frac{1}{a - x} + \frac{1}{b + x}}\right\} $$

$$ = \frac{1}{\sqrt{5}a}\mathop{\sum }\limits_{{n = 0}}^{\infty }{\left( \frac{x}{a}\right) }^{n} + \frac{1}{\sqrt{5}b}\mathop{\sum }\limits_{{n = 0}}^{\infty }{\left( -\frac{x}{b}\right) }^{n} $$

$$ = \mathop{\sum }\limits_{{n = 0}}^{\infty }\left\{ {\frac{1}{\sqrt{5}}\frac{1}{{a}^{n + 1}} + \frac{{\left( -1\right) }^{n}}{\sqrt{5}{b}^{n + 1}}}\right\} {x}^{n}\;\left( {\left| x\right| < a}\right) . $$

由此得到

$$ \frac{{f}^{\left( n\right) }\left( 0\right) }{n!} = \frac{1}{\sqrt{5}}\frac{1}{{a}^{n + 1}} + \frac{{\left( -1\right) }^{n}}{\sqrt{5}{b}^{n + 1}}. $$

因为 ${ab} = 1$ ,所以

$$ \frac{{f}^{\left( n\right) }\left( 0\right) }{n!} = \frac{1}{\sqrt{5}}{b}^{n + 1} + \frac{1}{\sqrt{5}}{\left( -1\right) }^{n}{a}^{n + 1} $$

$$ = \frac{{b}^{n + 1}}{\sqrt{5}}\left\lbrack {1 + {\left( -1\right) }^{n}{\left( \frac{a}{b}\right) }^{n + 1}}\right\rbrack $$

$$ \geq \frac{{b}^{n + 1}}{\sqrt{5}}\left\lbrack {1 - {\left( \frac{a}{b}\right) }^{n + 1}}\right\rbrack > 0. $$

又

$$ \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{\frac{n!}{{f}^{\left( n\right) }\left( 0\right) }}{{a}^{n + 1}} = \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{\sqrt{5}}{1 + {\left( -1\right) }^{n}{\left( \frac{a}{b}\right) }^{n + 1}} = \sqrt{5}, $$

而级数 $\displaystyle{\mathop{\sum }\limits_{{n = 0}}^{\infty }\sqrt{5}{a}^{n + 1}}$ 收敛,故 $\mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{n!}{{f}^{\left( n\right) }\left( 0\right) }$ 收敛.

证法 2 将 $f\left( x\right)$ 的表达式改写成方程:

$$ f\left( x\right) - {xf}\left( x\right) - {x}^{2}f\left( x\right) = 1. $$

两边求 $n$ 阶导数并代入 $x = 0$ 得到

$$ {f}^{\left( n\right) }\left( 0\right) - n{f}^{\left( n - 1\right) }\left( 0\right) - n\left( {n - 1}\right) {f}^{\left( n - 2\right) }\left( 0\right) = 0. $$

再两边除以 $n!$ ,推出

$$ \frac{{f}^{\left( n\right) }\left( 0\right) }{n!} = \frac{{f}^{\left( n - 1\right) }\left( 0\right) }{\left( {n - 1}\right) !} + \frac{{f}^{\left( n - 2\right) }\left( 0\right) }{\left( {n - 2}\right) !}. $$

又 $f\left( 0\right) = 1,{f}^{\prime }\left( 0\right) = 1$ ,令 ${a}_{n} = \frac{{f}^{\left( n\right) }\left( 0\right) }{n!}$ ,则有

$$ {a}_{n + 1} = {a}_{n} + {a}_{n - 1},\;{a}_{0} = {a}_{1} = 1. \tag{7.21} $$

下面只要证明 $\displaystyle{\mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{1}{{a}_{n}}}$ 收敛. 由 (7.21) 式推出

$$ {a}_{n + 1} = {a}_{n} + {a}_{n - 1} = 2{a}_{n - 1} + {a}_{n - 2} > 2{a}_{n - 1}, $$

于是

$$ \frac{\frac{1}{{a}_{{2k} + 1}}}{\frac{1}{{a}_{{2k} - 1}}} = \frac{{a}_{{2k} - 1}}{{a}_{{2k} + 1}} < \frac{1}{2},\;\frac{\frac{1}{{a}_{2k}}}{\frac{1}{{a}_{{2k} - 2}}} = \frac{{a}_{{2k} - 2}}{{a}_{2k}} < \frac{1}{2}, $$

由此得出级数 $\displaystyle{\mathop{\sum }\limits_{{k = 0}}^{\infty }\frac{1}{{a}_{{2k} + 1}}}$ 与级数 $\displaystyle{\mathop{\sum }\limits_{{k = 0}}^{\infty }\frac{1}{{a}_{2k}}}$ 收敛,从而

$$ \mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{1}{{a}_{n}} = \mathop{\sum }\limits_{{k = 0}}^{\infty }\left( {\frac{1}{{a}_{2k}} + \frac{1}{{a}_{{2k} + 1}}}\right) $$

收敛.

证法 3 设 $f\left( x\right) = \mathop{\sum }\limits_{{k = 0}}^{\infty }{a}_{k}{x}^{k}$ ,则 $\left( {1 - x - {x}^{2}}\right) \mathop{\sum }\limits_{{k = 0}}^{\infty }{a}_{k}{x}^{k} = 1$ . 整理后得

$$ {a}_{0} + \left( {{a}_{1} - {a}_{0}}\right) x + \mathop{\sum }\limits_{{m = 0}}^{\infty }\left( {{a}_{m + 2} - {a}_{m + 1} - {a}_{m}}\right) {x}^{m + 2} = 1. $$

由此推出 ${a}_{0} = 1,{a}_{1} = 1,{a}_{m + 2} = {a}_{m + 1} + {a}_{m}\left( {m = 0,1,\cdots }\right)$ . 下同证法 2.

证法 4 设 $a = \frac{\sqrt{5} - 1}{2},b = \frac{\sqrt{5} + 1}{2}$ ,则

$$ f\left( x\right) = \frac{1}{\sqrt{5}}\left\{ {\frac{1}{a - x} + \frac{1}{b + x}}\right\} , $$

$$ {f}^{\left( n\right) }\left( x\right) = \frac{1}{\sqrt{5}}\left\{ {\frac{n!}{{\left( a - x\right) }^{n + 1}} + \frac{{\left( -1\right) }^{n}n!}{{\left( x + b\right) }^{n + 1}}}\right\} , $$

由此推出 $\frac{{f}^{\left( n\right) }\left( 0\right) }{n!} = \frac{1}{\sqrt{5}{a}^{n + 1}} + \frac{{\left( -1\right) }^{n}}{\sqrt{5}{b}^{n + 1}}$ . 下同证法 1. 证法 ${5f}\left( x\right) = \frac{1}{1 - x\left( {1 + x}\right) } = \mathop{\sum }\limits_{{n = 0}}^{\infty }{x}^{n}{\left( 1 + x\right) }^{n}\left( {\left| {x\left( {1 + x}\right) }\right| < 1}\right)$ .

又 ${\left( 1 + x\right) }^{n} = \mathop{\sum }\limits_{{k = 0}}^{n}{C}_{n}^{k}{x}^{k}$ ,代入上式整理后得到

$$ f\left( x\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }\mathop{\sum }\limits_{{k = 0}}^{n}{C}_{n}^{k}{x}^{n + k} = \mathop{\sum }\limits_{{n = 0}}^{\infty }\mathop{\sum }\limits_{{m = n}}^{{2n}}{C}_{n}^{m - n}{x}^{m} = \mathop{\sum }\limits_{{m = 0}}^{\infty }\left( {\mathop{\sum }\limits_{{n \geq \frac{m}{2}}}^{m}{C}_{n}^{m - n}}\right) {x}^{m}. $$

由此得到

$$ \frac{{f}^{\left( m\right) }\left( 0\right) }{m!} = \mathop{\sum }\limits_{{n \geq \frac{m}{2}}}^{m}{C}_{n}^{m - n} $$

当 $m \geq 4$ 时,取 ${n}_{0} = m - 2 \geq \frac{m}{2}$ ,那么

$$ \mathop{\sum }\limits_{{n \geq \frac{m}{2}}}^{m}{C}_{n}^{m - n} > {C}_{{n}_{0}}^{m - {n}_{0}} = {C}_{m - 2}^{2} = \frac{\left( {m - 2}\right) \left( {m - 3}\right) }{2}. $$

从而

$$ \frac{m!}{{f}^{\left( m\right) }\left( 0\right) } < \frac{2}{\left( {m - 2}\right) \left( {m - 3}\right) } \leq \frac{16}{{m}^{2}}\;\left( {m \geq 4}\right) . $$

于是 $\mathop{\sum }\limits_{{m = 4}}^{\infty }\frac{m!}{{f}^{\left( m\right) }\left( 0\right) }$ 收敛,即有 $\mathop{\sum }\limits_{{m = 0}}^{\infty }\frac{m!}{{f}^{\left( m\right) }\left( 0\right) }$ 收敛.