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Referat Stoltz-Cesaro

Stoltz-Cesaro

Categorie: Referate Matematica

Referat downloadat de: 113 ori.

Cauta referat dupa: stoltz-cesaro.stoltz cesaro

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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 : $\displaystyle 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 : $\displaystyle (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 : $\displaystyle (l-\epsilon)\sum_{i=N(\epsilon)}^{k}(b_{i+1}-b_i) < \sum_{i=N(\ep... ...}(a_{n+1}-a_n) < (l+\epsilon)\sum_{i=N(\epsilon)}^{k}(b_{i+1}-b_i) \Rightarrow $ $\displaystyle (l-\epsilon)(b_{k+1}-b_{N(\epsilon)}) < a_{k+1} - a_{N(\epsilon)} < (l+\epsilon)(b_{k+1}-b_{N(\epsilon)})$ Divide the last relation by $ b_{k+1}>0$ to get : $\displaystyle (l-\epsilon)(1 - \frac{b_{N(\epsilon)}}{b_{k+1}}) < \frac{a_{k+1}... ...n)}}{b_{k+1}}<(l+\epsilon)(1 - \frac{b_{N(\epsilon)}}{b_{k+1}}) \Leftrightarrow$ $\displaystyle (l-\epsilon)(1 - \frac{b_{N(\epsilon)}}{b_{k+1}}) + \frac{a_{N(\e... ...psilon)(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 : $\displaystyle (l-\epsilon)<\frac{a_{k+1}}{b_{k+1}}< (l+\epsilon)$ (since the other terms who were left out converge to 0) This obviously means that : $\displaystyle \lim_{n \rightarrow \infty} \frac{a_n}{b_n}=l$ and we are done .

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