13
$\begingroup$

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

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

  • 0
    @RobertIsrael But you see I want to show that that expression tends to one. If it is not possible, let me know.2012-02-14

3 Answers 3

6

There is a proof at planetmath.org.

  • 2
    Please try to describe as much here as possible in order to make the answer self-contained. Links are fine as support, but they can go stale and then an answer which is nothing more than a link loses its value. Please read [this post](http://meta.stackexchange.com/questions/225370/your-answer-is-in-another-castle-when-is-an-answer-not-an-answer).2014-12-19
15

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.

  • 1
    @Peter: Graphics? I'm afraid that's too much work... Just draw your own picture by hand. It's not exactly difficult: a point, two lines through that point, two other lines through the origin, done. And notice that $(a_{n+1}-a_n)/(b_{n+1}-b_n)$ is the slope of the line segment between two consecutive points $(x,y)=(b_n,a_n)$; since all these segments have slopes between $\ell - \epsilon$ and $\ell + \epsilon$, the whole train of such segments starting at $(b_N,a_N)$ never goes outside of the wedge.2012-02-17
14

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

  • 0
    Thanks, Pedro, and thanks for the proof as well as the corollaries.2014-11-15