# 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}}

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})

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)})

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 ProofOfStolzCesaroTheorem 2013-03-22 13:17:45 2013-03-22 13:17:45 slash (33) slash (33) 4 slash (33) Proof msc 40A05