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Assume that $a_n\to \ell $ is a convergent sequence of complex numbers and $\{\lambda_n\}$ is a sequence of positive real numbers such that $\sum\limits_{k=0}^{\infty}\lambda_k = \infty$

Then, show that, $$\lim_{n\to\infty} \frac{1}{\sum_\limits{k=0}^{n}\lambda_k} \sum_\limits{k=0}^{n}\lambda_k a_k=\ell =\lim_{n\to\infty} a_n$$

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

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  • $\begingroup$ Is there any other restriction? With $\lambda_{k}=k$, the sum diverges (unless perhaps $l=0$). $\endgroup$ – Harry Sep 22 '17 at 12:22
  • $\begingroup$ there is no problem if the sum diverge. the result is still true see here for instance math.stackexchange.com/questions/2440315/… $\endgroup$ – Guy Fsone Sep 22 '17 at 12:24
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First you can assume that the $(a_n)$ are real, by considering the sequences $(\operatorname{Re} a_n)_n$ and $(\operatorname{Im} a_n)_n$ separately.

And then it is an immediate application of the Stolz–Cesàro theorem to $$ A_n := \sum_{k=0}^n \lambda_n a_n \quad, \quad B_n := \sum_{k=0}^n \lambda_n $$ since $(B_n)$ is strictly increasing and unbounded, and $$ \frac{A_{n+1} - A_n}{B_{n+1} - B_n} = a_{n+1} \to l $$

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Let $\varepsilon >0$ and $N$such 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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  • $\begingroup$ In the last inequality you mean to say the right-hand side goes to $\varepsilon$ as $n\to\infty$. $\endgroup$ – Blackbird Sep 22 '17 at 13:28
  • $\begingroup$ yes of course $\varepsilon $is negligible $\endgroup$ – Guy Fsone Sep 22 '17 at 13:35

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