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# 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 :

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

This obviously means that :

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

and we are done .

Keywords:

sequence,limit,convergence

Major Section:

Reference

Type of Math Object:

Proof

Parent:

## Mathematics Subject Classification

40A05*no label found*

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