Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

So, here we go again, the sequence $x_n$ is increasing and $x_n\to\infty$ as $n\to\infty$, and also, $\lim\limits_{n\to\infty}\dfrac{x_{n+1}}{x_n}= x$ which is a real non zero number,

Prove that :

$$\lim\limits_{n\to\infty}\frac{x_1+\cdots+x_{n+1}}{x_1+\cdots+x_n} = x $$

I'm stuck again, I know why it says that [eventually] $x_n>A$ for every $A$, from that I got :

$$\frac{x_1+\cdots+x_{n+1}}{x_1+\cdots+x_n} -x\leq (n+1)\frac{x_{n+1}}{x_n} -x$$ for every $n>N$

$N$ is special though but I don't get anywhere from there so it doesn't matter. can anybody help?

share|cite|improve this question
doesnt anybody have any ideas? im really stuck with this one – Plom Jul 24 '13 at 1:50
up vote 11 down vote accepted

Stolz-Cesaro says

Suppose $b_n$ is a sequence of strictly increasing numbers such that $b_n\nearrow +\infty$ and let $a_n$ be any sequence of real numbers. If $$\frac{a_{n+1}-a_n}{b_{n+1}-b_n}\to\ell $$ then $$\frac{a_n}{b_n}\to\ell$$

Now, let $$a_n=\sum_{k=1}^{n} x_k$$ $$b_n=\sum_{k=1}^{n-1} x_k$$

Then $$a_{n+1}-a_n=x_{n+1}\\ b_{n+1}-b_n=x_{n}$$

and $b_n\nearrow +\infty$, $b_n$ is strictly increasing.

For a proof, see here.

share|cite|improve this answer
I also have a proof here:… – marty cohen Jul 24 '13 at 4:20
@Peter Tamaroff the theorem is not exactly the way you state it. – user 1618033 Jul 24 '13 at 17:00
@Chris'ssis There are many versions. – Pedro Tamaroff Jul 24 '13 at 17:13
Two versions of Stolz-Cesaro theorem are compared in this answer. – Martin Sleziak Jul 24 '13 at 17:21
@MartinSleziak Neat. – Pedro Tamaroff Jul 24 '13 at 17:44

Your Answer


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.