回复 19#hbghlyj 这份讲义的60页的证明:
Theorem 2.3.11 (O.Stolz) Suppose $\left(x_{n}\right)$ and $\left(y_{n}\right)$ are two sequences of real numbers such that
(i) $y_{n} \rightarrow \infty$ as $n \rightarrow \infty$,
(ii) $\left(y_{n}\right)$ is a strictly increasing sequence (for large $\left.n\right)$, and
(iii) the limit
$$
\lim _{n \rightarrow \infty} \frac{x_{n}-x_{n-1}}{y_{n}-y_{n-1}}
$$
exists or tends to $\infty$ or $-\infty$. Then
$$
\lim _{n \rightarrow \infty} \frac{x_{n}}{y_{n}}=\lim _{n \rightarrow \infty} \frac{x_{n}-x_{n-1}}{y_{n}-y_{n-1}}
$$
Proof. The proof is similar to the proof of Theorem 2.3.6. Consider the case that $l=\lim _{n \rightarrow \infty} \frac{x_{n}-x_{n-1}}{y_{n}-y_{n-1}}$ is a number. Then for every $\varepsilon>0$ there is $N$ such that for $n>N$ we have
$$
\left|\frac{x_{n}-x_{n-1}}{y_{n}-y_{n-1}}-l\right|<\frac{\varepsilon}{2} .
$$
Since $\left(y_{n}\right)$ is strictly increasing eventually, so we can choose $N$ big enough so that $y_{k}-$ $y_{k-1}>0$ for all $k>N$ and therefore
$$
-\frac{\varepsilon}{2}\left(y_{k}-y_{k-1}\right)<x_{k}-y_{k-1}-l\left(y_{k}-y_{k-1}\right)<\frac{\varepsilon}{2}\left(y_{k}-y_{k-1}\right)
$$
Adding these inequalities over $k=N+1, \cdots, n$, where $n>N$, we obtain that
$$
-\frac{\varepsilon}{2}\left(y_{n}-y_{N}\right)<x_{n}-y_{N}-l\left(y_{n}-y_{N}\right)<\frac{\varepsilon}{2}\left(y_{n}-y_{N}\right)
$$
which can be written as, since $y_{n}-y_{N}>0$
$$
\left|\frac{x_{n}-x_{N}}{y_{n}-y_{N}}-l\right|<\frac{\varepsilon}{2}
$$
for all $n>N .$ Next we use the identity (similar to that in the proof of Theorem 2.3.6)
$$
\frac{x_{n}}{y_{n}}-l=\frac{x_{N}-l y_{N}}{y_{n}}+\left(1-\frac{y_{N}}{y_{n}}\right)\left(\frac{x_{n}-x_{N}}{y_{n}-y_{N}}-l\right)
$$
so that
$$
\left|\frac{x_{n}}{y_{n}}-l\right|<\left|\frac{x_{N}-l y_{N}}{y_{n}}\right|+\frac{\varepsilon}{2}
$$
for every $n>N$. Since $y_{n} \rightarrow \infty$ so that
$$
\frac{x_{N}-l y_{N}}{y_{n}} \rightarrow 0 \quad \text { as } n \rightarrow \infty
$$
Therefore there is $N_{1}>N$ such that
$$
\left|\frac{x_{N}-l y_{N}}{y_{n}}\right|<\frac{\varepsilon}{2} \quad \text { for } n>N_{1}
$$
and therefore
$$
\left|\frac{x_{n}}{y_{n}}-l\right|<\left|\frac{x_{N}-l y_{N}}{y_{n}}\right|+\frac{\varepsilon}{2}<\varepsilon
$$
for every $n>N_{1}$. By definition
$$
\lim _{n \rightarrow \infty} \frac{x_{n}}{y_{n}}=l=\lim _{n \rightarrow \infty} \frac{x_{n}-x_{n-1}}{y_{n}-y_{n-1}}
$$
and the proof is complete.$\blacksquare$
As as example, if $k$ is a positive integer, then we can show (Exercise) by Stolz's theorem that
$$
\lim _{n \rightarrow \infty} \frac{1^{k}+2^{k}+\cdots+n^{k}}{n^{k+1}}=\frac{1}{k+1}
$$