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

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

  • $\begingroup$ 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. $\endgroup$ Feb 14, 2012 at 16:43
  • $\begingroup$ @RobertIsrael But you see I want to show that that expression tends to one. If it is not possible, let me know. $\endgroup$
    – Pedro
    Feb 14, 2012 at 16:44

3 Answers 3


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.

  • $\begingroup$ Could you help me out with some graphics? Or a less intuitive approach? Upvoted anyways. $\endgroup$
    – Pedro
    Feb 16, 2012 at 22:08
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    $\begingroup$ @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. $\endgroup$ Feb 17, 2012 at 6:56
  • $\begingroup$ How do you know, tht for sufficently big x, the point will not be with in the the triangle, of $y=(l\pm \epsilon) x$ but in $2\epsilon$ religion? $\endgroup$
    – M Desmond
    Oct 22, 2021 at 16:13
  • $\begingroup$ @MDesmond: I'm not sure I understand your question. (Even if I replace the word “religion” with “region”!) Could you try to be more precise? $\endgroup$ Oct 22, 2021 at 17:04
  • $\begingroup$ "$y = (\ell - \epsilon) x$ 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 narrower wedge." So whats wrong with that? Why again consider $(\ell +2\epsilon , \ell+2 \epsilon) $ ? $\endgroup$
    – M Desmond
    Oct 22, 2021 at 18:14

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

  • $\begingroup$ just wondering if lim am/bm = $\alpha$, shouldn't $\lim \frac{a_{n+1} - a_n}{b_{n+1}-b_n}$ > $\lim \frac{a_m}{b_m}$ $\endgroup$ Apr 21, 2014 at 8:29
  • $\begingroup$ referring to the line before the first corollary. I don't really understand how the conclusion is arrived $\endgroup$ Apr 21, 2014 at 8:31
  • $\begingroup$ Are you assuming in this proof that $(b_n)$ diverges to infinity? $\endgroup$
    – user84413
    Nov 15, 2014 at 0:27
  • $\begingroup$ @user84413 I have clarified it. $\endgroup$
    – Pedro
    Nov 15, 2014 at 3:56
  • $\begingroup$ Thanks, Pedro, and thanks for the proof as well as the corollaries. $\endgroup$
    – user84413
    Nov 15, 2014 at 18:00

There is a proof at planetmath.org.

  • 5
    $\begingroup$ @Peter: Could you edit your question to explain that you're looking for an approach other than the one given there? $\endgroup$ Feb 14, 2012 at 10:35
  • $\begingroup$ @ZevChonoles Yes, no problem. $\endgroup$
    – Pedro
    Feb 14, 2012 at 17:06
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    $\begingroup$ 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. $\endgroup$ Feb 14, 2012 at 18:06
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    $\begingroup$ @HansLundmark Can you answer with a corrected proof? $\endgroup$
    – Pedro
    Feb 15, 2012 at 0:14
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    $\begingroup$ 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. $\endgroup$
    – robjohn
    Dec 19, 2014 at 22:20

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