Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question
    
Why did this get rated down? Is it because of the formatting? –  Tyler Feb 14 '12 at 16:24
    
@TylerBailey I don't know. I think the formatting is good. I'd like votes not to be anonymous. –  Pedro Tamaroff Feb 14 '12 at 16:29
    
Since $\displaystyle \frac{\frac{a_{n+1}}{a_n} - 1}{\frac{b_{n+1}}{b_n} - 1} = \frac{b_n}{a_n} \frac{a_{n+1} - a_n}{b_{n+1}-b_n}$, this is just restating the original statement, at least if $\ell_1 \ne 0$. You'd also have to treat the case $\ell_1 = 0$ separately. –  Robert Israel Feb 14 '12 at 16:43
    
@RobertIsrael But you see I want to show that that expression tends to one. If it is not possible, let me know. –  Pedro Tamaroff Feb 14 '12 at 16:44

3 Answers 3

up vote 4 down vote accepted

There is a proof here.

share|improve this answer
3  
@Peter: Could you edit your question to explain that you're looking for an approach other than the one given there? –  Zev Chonoles Feb 14 '12 at 10:35
    
@ZevChonoles Yes, no problem. –  Pedro Tamaroff Feb 14 '12 at 17:06
    
It seems to me that the final step in that proof is not quite right. But if one writes $$l-2\epsilon < \frac{a_{k+1}}{b_{k+1}} < l+2\epsilon$$ instead, then it should be OK. –  Hans Lundmark Feb 14 '12 at 18:06
    
@HansLundmark Can you answer with a corrected proof? –  Pedro Tamaroff Feb 15 '12 at 0:14
    
@Peter: I though I just did... ;-) But since you asked for it, I've written a separate answer. –  Hans Lundmark Feb 15 '12 at 9:06

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.

share|improve this answer
    
Could you help me out with some graphics? Or a less intuitive approach? Upvoted anyways. –  Pedro Tamaroff Feb 16 '12 at 22:08
    
@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. –  Hans Lundmark Feb 17 '12 at 6:56

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

share|improve this answer
    
just wondering if lim am/bm = $\alpha$, shouldn't $\lim \frac{a_{n+1} - a_n}{b_{n+1}-b_n}$ > $\lim \frac{a_m}{b_m}$ –  user136266 Apr 21 at 8:29
    
referring to the line before the first corollary. I don't really understand how the conclusion is arrived –  user136266 Apr 21 at 8:31

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.