每日一题[862]舍小利以谋远

定义在${\bf{R}}$上的函数$f\left( x \right) = \dfrac{{{4^x}}}{{{4^x} + 2}}$，

$${S_n} = f\left( {\dfrac{1}{n}} \right) + f\left( {\dfrac{2}{n}} \right) + \cdots + f\left( {\dfrac{{n - 1}}{n}} \right),n = 2,3, \cdots .$$ (1) 求${S_n}$； 

(2) 是否存在常数$M > 0$，$\forall n \geqslant 2$，有

$$\dfrac{1}{{{S_2}}} + \dfrac{1}{{{S_3}}} + \cdots + \dfrac{1}{{{S_{n + 1}}}} \leqslant M.$$

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分析与解　(1) 注意到

$$\begin{split}f\left( x \right) + f\left( {1 - x} \right) &= \dfrac{{{4^x}}}{{{4^x} + 2}} + \dfrac{{{4^{1 - x}}}}{{{4^{1 - x}} + 2}} \\&= \dfrac{{{4^x}}}{{{4^x} + 2}} + \dfrac{4}{{4 + 2 \cdot {4^x}}} = 1.\end{split}$$ 所以 $$\begin{split}2{S_n} &= \left[ {f\left( {\dfrac{1}{n}} \right) + f\left( {\dfrac{{n - 1}}{n}} \right)} \right] + \left[ {f\left( {\dfrac{2}{n}} \right) + f\left( {\dfrac{{n - 2}}{n}} \right)} \right] + \cdots + \left[ {f\left( {\dfrac{{n - 1}}{n}} \right) + f\left( {\dfrac{1}{n}} \right)} \right]\\& = n - 1.\end{split}$$ 所以${S_n} = \dfrac{{n - 1}}{2}$．

(2) 因为

$$\dfrac{1}{{{S_2}}} + \dfrac{1}{{{S_3}}} + \cdots + \dfrac{1}{{{S_{n + 1}}}} = \dfrac{2}{1} + \dfrac{2}{2} + \cdots + \dfrac{2}{n}.$$ 构造数列$\left\{ {{b_n}} \right\}$：$\dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{4},\dfrac{1}{8},\dfrac{1}{8}, \dfrac{1}{8} , \dfrac{1}{8}, \cdots$（其中形如$\dfrac{1}{{{2^k}}}$的项有${2^{k - 1}}$ 个）截止到最后一个$\dfrac{1}{{{2^k}}}$，共有 $$1 + 2 + {2^2} + \cdots + {2^{k - 1}} = {2^k} - 1$$ 项．所以取$n = {2^k}$，则 $$\begin{split}\dfrac 12\sum\limits_{k=2}^{n+1}{\dfrac 1{S_{n + 1}}} &= 1 + \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{n}\\& = 1 + \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{2^{k}}\\& > 1 + \dfrac{1}{2} + \underbrace {\dfrac{1}{4} + \dfrac{1}{4}}_2 + \underbrace {\dfrac{1}{8} + \dfrac{1}{8} + \cdots + \dfrac{1}{8}}_4 + \cdots + \underbrace {\dfrac{1}{{{2^k}}} + \dfrac{1}{{{2^k}}} + \cdots \dfrac{1}{{{2^k}}}}_{{2^{k - 1}}}\\& = 1 + \dfrac{k}{2}.\end{split}$$ 因此不存在常数$M > 0$，$\forall n \geqslant 2$，有 $$\dfrac{1}{{{S_2}}} + \dfrac{1}{{{S_3}}} + \cdots + \dfrac{1}{{{S_{n + 1}}}} \leqslant M.$$