2.4 Stolz定理

对于复杂的分式形式的求极限问题, Stolz 定理是一个强有力的工具.

Stolz 定理

设 ${y_{n}}$ 严格单调递增, 且 $lim_{n \to \infty} y_{n} = + \infty$ , 且 $lim_{n \to \infty} \frac{x_{n} - x_{n - 1}}{y_{n} - y_{n - 1}} = A$ , 则 $lim_{n \to \infty} \frac{x_{n}}{y_{n}} = A$ . 这里 $A \in R$ , 或 $A = \pm \infty$ .

证明

情形一: $A \in R$ .
由条件可令 $\frac{x_{n} - x_{n - 1}}{y_{n} - y_{n - 1}} = A + α_{n} (α_{n} = o (1))$ . 则 $\forall ε > 0, \exists N_{1} \in N^{*}, \forall n > N_{1} : | α_{n} | < \frac{ε}{2}$ . 而由假设得

x_{n} - x_{n - 1} = A (y_{n} - y_{n - 1}) + α_{n} (y_{n} - y_{n - 1}) .

累加得

x_{n} - x_{N_{1}} = \sum_{i = N_{1} + 1}^{n} (x_{i} - x_{i - 1}) = A (y_{n} - y_{N_{1}}) + \sum_{i = N_{1} + 1}^{n} α_{i} (y_{i} - y_{i - 1}) .

从而

\begin{aligned} | \frac{x_{n}}{y_{n}} - A | & = \frac{| x_{n} - A y_{n} |}{| y_{n} |} = \frac{| x_{n} - x_{N_{1}} + x_{N_{1}} - A y_{n} |}{| y_{n} |} \\ \leq \frac{| x_{N_{1}} - A y_{N_{1}} |}{| y_{n} |} + \frac{| \sum_{i = N_{1} + 1}^{n} α_{i} (y_{i} - y_{i - 1}) |}{| y_{n} |} . \end{aligned}

又 $lim_{n \to \infty} \frac{| x_{N_{1}} - A y_{N_{1}} |}{| y_{n} |} = 0$ , 故 $\exists N_{2} \in N^{*}, \forall n > N_{2} : | \frac{x_{N_{1}} - A y_{N_{1}}}{y_{n}} | < \frac{ε}{2}$ . 于是 $\forall n > max {N_{1}, N_{2}}$ ,

| \frac{x_{n}}{y_{n}} - A | < \frac{ε}{2} + \frac{\frac{ε}{2}}{y_{n}} \sum_{i = N_{1} + 1}^{n} (y_{i} - y_{i - 1}) = \frac{ε}{2} + \frac{ε}{2 y_{n}} (y_{n} - y_{N_{1}}) < \frac{ε}{2} + \frac{ε}{2} = ε .

从而 $lim_{n \to \infty} \frac{x_{n}}{y_{n}} = A$ , 结论获证.

情形二: $A = + \infty$ .
转化为情形一. 由已知, $lim_{n \to \infty} \frac{y_{n} - y_{n - 1}}{x_{n} - x_{n - 1}} = 0$ , 因此只要证当 $n$ 充分大时 ${x_{n}}$ 严格单调递增且趋于正无穷.
事实上, 由 $lim_{n \to \infty} \frac{x_{n} - x_{n - 1}}{y_{n} - y_{n - 1}} = + \infty$ , 则

\exists N_{2} \in N^{*}, \forall n \geq N_{2} : \frac{x_{n} - x_{n - 1}}{y_{n} - y_{n - 1}} > 1 \Rightarrow x_{n} - x_{n - 1} > y_{n} - y_{n - 1} > 0.

从而 ${x_{n}}$ 严格单调递增. 此时

x_{n} - x_{N_{2}} = \sum_{i = N_{2} + 1}^{n} (x_{i} - x_{i - 1}) > \sum_{i = N_{2} + 1}^{n} (y_{i} - y_{i - 1}) = y_{n} - y_{N_{2}} \to + \infty (n \to \infty) .

故 $lim_{n \to \infty} x_{n} = + \infty$ . 根据情形一的要求, 结论获证.

情形三: $A = - \infty$ .
则 $lim_{n \to \infty} \frac{- x_{n} - (- x_{n - 1})}{y_{n} - y_{n - 1}} = - lim_{n \to \infty} \frac{x_{n} - x_{n - 1}}{y_{n} - y_{n - 1}} = + \infty$ . 由情形二推得,

lim_{n \to \infty} \frac{- x_{n}}{y_{n}} = + \infty \Rightarrow lim_{n \to \infty} \frac{x_{n}}{y_{n}} = - \infty .

从而结论获证.

推论

若 $lim_{n \to \infty} x_{n} = a$ , 则 $lim_{n \to \infty} \frac{x_{1} + \dots + x_{n}}{n} = a$ . 这里 $a \in R$ 或 $a = \pm \infty$ .