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The Stolz–Cesàro theorem states that:

Let $a_n$ and $b_n$ be two sequences of real numbers such that:

$b_n$ is strictly monotone

$\lim_{n \to \infty} b_n=\infty$

$\lim_{n \to \infty} \dfrac{a_{n+1}-a_n}{b_{n+1}-b_n}=l$

Then, the limit $\lim_{n\to \infty}\dfrac{a_n}{b_n}$

also exists and equals $l$

I was looking for a proof of the following theorem and I came across this. But by following the same approach with a little manipulation(which is bound to be wrong)leads to the same result without imposing the necessary conditions $1)$ and $2)$ on $b_n$

Case 1

If $b_N=0$, draw two lines through the origin with slope $l+\epsilon$ and $l-\epsilon$, then $l+\epsilon<\dfrac{a_n}{b_n}<l+\epsilon$ for all $n\ge N$

Case 2:

If $b_n\neq 0$, then choose $n_1$ and $n_2$ such that;

$\dfrac{a_N}{b_N}<l+n_{1} \epsilon$

$\dfrac{a_N}{b_N}>l-n_{2}\epsilon$

Let $n=\max(n_1,n_2)$

Now,

$l-n\epsilon<\dfrac{a_n}{b_n}<n+l\epsilon$

Hence, $\lim_{n\to \infty}\dfrac{a_n}{b_n}=l$

Question:

Whats wrong with this and where did I make the blunder?

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  • First, if we're interested in $\lim_{n \rightarrow\infty}\frac{a_n}{b_n}$, we must assume that $b_n \neq 0$ eventually. So we can as well assume that $b_n \neq 0$ always holds. So we don't need your case 1. – jflipp Nov 13 '14 at 15:02
  • Second, I can't follow the reasoning in your case 2. Maybe it would help if you stated the preconditions for your claim of convergence more explicitly, and also if you formulated your proof of convergence in the language we're all used to, namely "pick an $\epsilon > 0$, find an $N \in \mathbb N$ such that $\forall n>N:|\frac{a_n}{b_n} - l| <\epsilon$". How exactly do you find that $N$? Why exactly does it have the desired properties? – jflipp Nov 13 '14 at 15:04

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