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 .