导入：
$\displaystyle 1. 已知 \lim_{n \to \infty }a_n \sum_{i=1}^na_i^2=1, 求证 \lim_{n \to \infty }\sqrt[3]{3n}a_n =1 .$
$\displaystyle \text{解: 设 } S_n= \sum_{i=1}^na_i^2$
$\displaystyle \text{则题设即为 } \lim_{n \to \infty }a_n S_n=1$
$\displaystyle \text{要想证 } \lim_{n \to \infty} \sqrt{3n}a_n=1$
$\displaystyle \text{只需证 } \lim_{n \to \infty} \frac{1}{\sqrt[3]{3n}a_n}=1$
$\displaystyle \Longleftarrow \lim_{n \to \infty} \frac{1}{3na_n^3}=1$
$\displaystyle \text{而 } \lim_{n \to \infty} (a_nS_n)^3=1, \text{ 则只需证 } \lim_{n \to \infty} \frac{S_n^3}{3n}=1$
$\displaystyle \text{而 } \lim_{n \to \infty}S_n=\lim_{n \to \infty}n=\infty, \text{那么只需证 } \lim_{n \to \infty} (S_n^3-S_{n-1}^3)=3$
$\displaystyle \Longleftarrow \lim_{n \to \infty}a_n^2(S_n^2+S_nS_{n-1}+S_{n-1}^2)=3$
$\displaystyle \Longleftarrow \lim_{n \to \infty}a_n^2S_n^2+\lim_{n \to \infty}a_n^2S_nS_{n-1}+\lim_{n \to \infty}a_n^2S_{n-1}^2=3$
$\displaystyle \text{由题设这是显然的，因此证毕.}$
感觉这道题主要是不会 $\operatorname{Stolz} $ 定理所以做不出来，但是似乎会了也做不出来。
必要性探路法还是得精进一下理解。
$\displaystyle1.1 ;(\operatorname{Stolz} 定理) 已知 {y_n} 严格单调递增,\lim_{n \to \infty}y_n=\infty,\lim_{n \to \infty} \frac{x_n-x_{n-1}}{y_n-y_{n-1}}存在$
$\displaystyle 则\lim_{n \to \infty} \frac{x_n}{y_n}=\lim_{n \to \infty} \frac{x_n-x_{n-1}}{y_n-y_{n-1}}.$
$\displaystyle \text{证明: 设 } \lim_{n \to \infty} \frac{x_n-x_{n-1}}{y_n-y_{n-1}}=a$
$\displaystyle \text{则存在 } N, \text{当 } n>N \text{ 时 } \left|\frac{x_n-x_{n-1}}{y_n-y_{n-1}}-a\right|< \frac{\varepsilon}{2} \text{ 且 } y_n>0$
$\displaystyle \text{即 } a-\frac{\varepsilon}{2}<\frac{x_{N+1}-x_{N}}{y_{N+1}-y_{N}}<a+\frac{\varepsilon}{2}$
$\displaystyle a-\frac{\varepsilon}{2}<\frac{x_{N+2}-x_{N+1}}{y_{N+2}-y_{N+1}}<a+\frac{\varepsilon}{2}$
$\displaystyle \dots$
$\displaystyle a-\frac{\varepsilon}{2}<\frac{x_{n}-x_{n-1}}{y_{n}-y_{n-1}}<a+\frac{\varepsilon}{2}$
$\displaystyle \text{将分母乘上去并求和得}$
$\displaystyle a-\frac{\varepsilon}{2}<\frac{x_{n}-x_{N+1}}{y_{n}-y_{N+1}}<a+\frac{\varepsilon}{2}$
$\displaystyle \text{即 } \left|\frac{x_n-x_{N+1}}{y_n-y_{N+1}}-a\right|< \frac{\varepsilon}{2}$
$\displaystyle \text{而显然 } \lim_{n \to \infty}\frac{x_{N+1}-ay_{N+1}}{y_n}=0$
$\displaystyle \text{故存在 } N_1, \text{当 } n>N_1 \text{ 时 } \left|\frac{x_{N+1}-ay_{N+1}}{y_n}\right|<\frac{\varepsilon}{2}$
$\displaystyle \text{取 } N_2=\max{N_1,N} \text{ 则当 } n>N_2 \text{ 时}$
$\displaystyle \left|\frac{x_n}{y_n}-a\right|=\left|\frac{x_n-ay_n}{y_n}\right|$
$\displaystyle =\left|\frac{x_n-ay_n+x_{N+1}+ay_{N+1}-x_{N+1}-ay_{N+1}}{y_n}\right|$
$\displaystyle =\left|\frac{x_{N+1}-ay_{N+1}}{y_n}+\frac{x_n-x_{N+1}-(ay_n-ay_{N+1})}{y_n}\right|$
$\displaystyle =\left|\frac{x_{N+1}-ay_{N+1}}{y_n}+\frac{x_n-x_{N+1}-(ay_n-ay_{N+1})}{y_n-y_{N+1}} \times \left(1-\frac{y_{N+1}}{y_n}\right)\right|$
$\displaystyle \leq \left|\frac{x_{N+1}-ay_{N+1}}{y_n}\right|+\left|\frac{x_n-x_{N+1}}{y_n-y_{N+1}}-a\right| \leq \varepsilon$
$\displaystyle \text{由此命题得证}$