Stolz-Cesàro Theorem

Question

Recently I've been trying to find a satisfactory proof of the Stolz-Cesàro Theorem but I havent found any. As I remember the claim is as follows:

Let $${\left\{ {{b_n}} \right\}_{n \in {\Bbb N}}}$$ be a sequence such that

$${b_{k + 1}} - {b_k} > 0 $$ and $$ \mathop {\lim }\limits_{k \to \infty }\sum_{n=0}^{k} {b_n} = \infty $$

Then if $${\left\{ {{a_n}} \right\}_{n \in {\Bbb N}}}$$ is another sequence and the limit

$$\mathop {\lim }\limits_{n \to \infty } \frac{{{a_{n + 1}} - {a_n}}}{{{b_{n + 1}} - {b_n}}} = \ell_1 $$

exists, then

$$\mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}} = {\ell _2}$$

exists too and

$${\ell _1} = {\ell _2}$$

Answer 1

I find it easiest to view this geometrically. With $\ell - \epsilon < \frac{{{a_{n + 1}} - {a_n}}}{{{b_{n + 1}} - {b_n}}} < \ell + \epsilon$ for $n \ge N$, all the points $(x,y)=(b_n,a_n)$ for $n \ge N$ will lie inside the wedge formed by the two lines through the point $(x,y)=(b_N,a_N)$ with slopes $\ell - \epsilon$ and $\ell + \epsilon$, respectively. And this wedge will, for large enough $x$, stay entirely within the wider wedge formed by the lines $y = (\ell - 2 \epsilon) x$ and $y = (\ell + 2 \epsilon) x$ through the origin. (This step is where the PlanetMath proof is not quite precise; the statement is not necessarily true if you take the lines $y = (\ell - \epsilon) x$ and $y = (\ell + \epsilon) x$.) Since $b_n \nearrow +\infty$, all points $(x,y)=(b_n,a_n)$ for $n \ge M$, say, will have large enough $x$ coordinate to lie in the part of the narrower wedge that lies inside the wider wedge; thus $\ell - 2 \epsilon < \frac{a_n}{b_n} < \ell + 2 \epsilon$ for $n \ge M$. Done.

Answer 2

Here's a more general situation:

THM Let $\langle a_n\rangle$ be any sequence of real numbers and suppose that $\langle b_n\rangle $ is a sequence of positive numbers such that $b_n$ is strictly monotone increasing to $\infty$. Then $$\liminf_{n\to\infty}\frac{a_n}{b_n}\geq \liminf_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$ $$\limsup_{n\to\infty}\frac{a_n}{b_n}\leq \limsup_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$

PROOF We prove the case for $\liminf$; the $\limsup$ case is analogous. Take $$\alpha <\liminf_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$

Then there exists $N$ such that for each $k\geq 0$ we have $$\alpha <\frac{a_{N+k}-a_{N+k-1}}{b_{N+k}-b_{N+k-1}}$$ Since $b_{n+1}>b_n$, we have for $k\geq 0$ that $$\alpha \left( {{b_{N + k}} - {b_{N + k - 1}}} \right) < {a_{N + k}} - {a_{N + k - 1}}$$

Thus, for any $m\geq 0$, $$\eqalign{ \alpha \sum\limits_{k = 0}^m {\left( {{b_{N + k}} - {b_{N + k - 1}}} \right)} & < \sum\limits_{k = 0}^m {\left( {{a_{N + k}} - {a_{N + k - 1}}} \right)} \cr \alpha \left( {{b_{N + m}} - {b_{N - 1}}} \right) &< {a_{N + m}} - {a_{N - 1}} \cr} $$

It follows that $$\alpha \left( {1 - \frac{{{b_{N - 1}}}}{{{b_{N + m}}}}} \right) < \frac{{{a_{N + m}}}}{{{b_{N + m}}}} - \frac{{{a_{N - 1}}}}{{{b_{N + m}}}}$$ and taking $m\to\infty$ $$\alpha \leq \mathop {\lim \inf }\limits_{m \to \infty } \frac{{{a_m}}}{{{b_m}}}$$

It follows that, for each $\alpha <\liminf\limits_{n\to\infty}\dfrac{a_{n+1}-a_n}{b_{n+1}-b_n}$ we have $\alpha \leq \liminf\limits_{m \to \infty } \dfrac{{{a_m}}}{{{b_m}}}$, which means $$\mathop {\liminf }\limits_{n \to \infty } \dfrac{{{a_{n + 1}} - {a_n}}}{{{b_{n + 1}} - {b_n}}} \leq \liminf\limits_{m\to\infty} \frac{{{a_m}}}{{{b_m}}}$$

COR Let $\langle a_n\rangle$ and $\langle b_n\rangle$ be as before. Then if $$\ell=\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$ exists, so does $$\ell'=\lim_{n\to\infty}\frac{a_n}{b_n}$$ and $\ell=\ell'$

COR Let $x_n$ be any sequence. If $$\lim_{n\to\infty} x_n=\ell$$ then $$\lim_{n\to\infty}\frac 1 n \sum_{k=1}^n x_k=\ell$$

P By the first corollary with $b_n=n$ and $a_n=\sum_{k=1}^n x_k$, we have $$\mathop {\lim }\limits_{n \to \infty } \frac{{\sum\limits_{k = 1}^{n + 1} {{a_k}} - \sum\limits_{k = 1}^n {{a_k}} }}{{n + 1 - n}} = \mathop {\lim }\limits_{n \to \infty } {x_{n + 1}} = \mathop {\lim }\limits_{n \to \infty } {x_n}$$

which means $$\mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {{x_k}} = \mathop {\lim }\limits_{n \to \infty } {x_n}$$

Answer 3

There is a proof at planetmath.org.