Sunday, 18 November 2018

calculus - General Cesaro summation with weight




Assume that an is a convergent sequence of complex numbers and {λn} is a sequence of positive real numbers such that k=0λk=





Then, show that,
lim



(Note that : This is more general than the special case where, \lambda_n= 1)



Answer



Let \varepsilon >0 and Nsuch that |a_k-l|\le \varepsilon for all k>N
Then, for n>N we have,

\begin{split}\left| \frac{\sum_\limits{k=0}^{n}\lambda_k a_k}{\sum_\limits{k=0}^{n}\lambda_k} -l\right| &= &\left| \frac{\sum_\limits{k=0}^{n}\lambda_k (a_k - l)}{\sum_\limits{k=0}^{n}\lambda_k} \right|\\ &= &\left| \frac{\sum_\limits{k=0}^{N}\lambda_k (a_k - l)+\sum_\limits{k=N}^{n}\lambda_k (a_k - l)}{\sum_\limits{k=0}^{n}\lambda_k} \right|\\ &\le & \frac{M}{\sum_\limits{k=0}^{n}\lambda_k} + \frac{\sum_\limits{k=N}^{n}\lambda_k \underbrace{\left| a_k - l\right|}_{\le\varepsilon}}{\sum_\limits{k=0}^{n}\lambda_k} \\ &\le& \frac{M}{\sum_\limits{k=0}^{n}\lambda_k} + \varepsilon\to 0 \end{split}
since \sum_\limits{k=0}^{N}\lambda_k\to \infty.
Where M= \left|\sum_\limits{k=0}^{N}\lambda_k( a_k-l)\right|


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