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Suppose that, $a_n\to \ell\neq 0$ is a converging sequence of non vanishing` complex numbers and $\{\lambda_n\}$ is a sequence of positifs real numbers such that $\sum\limits_{k=0}^{\infty}\lambda_k = \infty$

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

I have no clue on how to start.

Guy Fsone
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    Hint : write $\ell$ as $\frac{\displaystyle\sum_{k=0}^n \lambda_k}{\displaystyle\sum_{k=0}^n \frac{\lambda_k}{\ell}}$. Notice that if $u_n\neq 0, l\neq 0$, $u_n\to l$ if and only if $\frac{1}{u_n} \to \frac{1}{l}$. – Maxime Ramzi Sep 23 '17 at 13:24
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    This is a sort of generalized Cesàro mean, in disguise. Relabel $b_n:=a_n^{-1}$. Then $\lim_nb_n=\ell^{-1}$ and you seek to show that $$ {\sum_{k=1}^n \lambda_k b_k\over\sum_{k=1}^n\lambda_k}=\ell^{-1}. $$ The special case $\lambda_n=1$ is the familiar assertion that if a sequence converges in the usual sense, then it also converges in the sense of Cesàro convergence (that is, its running averages converges). You might try this special case as a warm-up -- the proof is a pretty straightforward exercise in epsilonics. – John Dawkins Sep 23 '17 at 15:58

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Using this : General Cesaro : $\lim_{n\to\infty} \frac{\sum_\limits{k=0}^{n}\lambda_k a_k}{\sum_\limits{k=0}^{n}\lambda_k} =\ell =\lim_{n\to\infty} a_n$ to the sequence, $\frac{1}{a_n}$

we have that $$\lim_{n\to\infty} \frac{\sum_\limits{k=0}^{n}\lambda_k a^{-1}_k}{\sum_\limits{k=0}^{n}\lambda_k} =\ell^{-1} =\lim_{n\to\infty}a^{-1}_n$$ that is $$\lim_{n\to\infty}\left(\sum_\limits{k=0}^{n}\lambda_k\right)\left(\sum_\limits{k=0}^{n}\frac{\lambda_k }{a_k}\right)^{-1}= \ell =\lim_{n\to\infty} a_n$$

Guy Fsone
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