Proof of Stolz-Cesaro theorem

From the definition of convergence , for every $\epsilon>0$ there is $N(\epsilon)\in\mathbb{N}$ such that $(\forall)n\geq N(\epsilon)$ , we have :

l-\epsilon<\frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}}<l+\epsilon

Because $b_{n}$ is strictly increasing we can multiply the last equation with $b_{n+1}-b_{n}$ to get :

(l-\epsilon)(b_{n+1}-b_{n})<a_{n+1}-a_{n}<(l+\epsilon)(b_{n+1}-b_{n})

Let $k>N(\epsilon)$ be a natural number . Summing the last relation we get :

(l-\epsilon)\sum_{i=N(\epsilon)}^{k}(b_{i+1}-b_{i})<\sum_{i=N(\epsilon)}^{k}(a% _{n+1}-a_{n})<(l+\epsilon)\sum_{i=N(\epsilon)}^{k}(b_{i+1}-b_{i})\Rightarrow

(l-\epsilon)(b_{k+1}-b_{N(\epsilon)})<a_{k+1}-a_{N(\epsilon)}<(l+\epsilon)(b_{% k+1}-b_{N(\epsilon)})

Divide the last relation by $b_{k+1}>0$ to get :

(l-\epsilon)(1-\frac{b_{N(\epsilon)}}{b_{k+1}})<\frac{a_{k+1}}{b_{k+1}}-\frac{% a_{N(\epsilon)}}{b_{k+1}}<(l+\epsilon)(1-\frac{b_{N(\epsilon)}}{b_{k+1}})\Leftrightarrow

(l-\epsilon)(1-\frac{b_{N(\epsilon)}}{b_{k+1}})+\frac{a_{N(\epsilon)}}{b_{k+1}% }<\frac{a_{k+1}}{b_{k+1}}<(l+\epsilon)(1-\frac{b_{N(\epsilon)}}{b_{k+1}})+% \frac{a_{N(\epsilon)}}{b_{k+1}}

This means that there is some $K$ such that for $k\geq K$ we have :

(l-\epsilon)<\frac{a_{k+1}}{b_{k+1}}<(l+\epsilon)

(since the other terms who were left out converge to 0)

This obviously means that :

\lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}}=l

and we are done .

Title	Proof of Stolz-Cesaro theorem
Canonical name	ProofOfStolzCesaroTheorem
Date of creation	2013-03-22 13:17:45
Last modified on	2013-03-22 13:17:45
Owner	slash (33)
Last modified by	slash (33)
Numerical id	4
Author	slash (33)
Entry type	Proof
Classification	msc 40A05