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It is well known that for any sequence $\{x_n\}$ of real or complex numbers which converges to a limit $x$, the sequence of averages of the first $n$ terms is also convergent to $x$. That is, the sequence $\{a_n\}$ defined by

$$a_n = \frac{x_1+x_2+\ldots + x_n}{n}$$

converges to $x$. How "severe" of a weighting function $w(n)$ can we create that the sequence of weighted averages $\{b_n\}$ defined by

$$b_n = \frac{w(1)x_1 + w(2)x_2 + \ldots + w(n)x_n}{w(1)+w(2)+\ldots+w(n)} $$

is convergent to $x$? Is it possible to choose $w(n)$ such that $\{b_n\}$ is divergent?

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I think you might want to ensure that $w(n)>0$ – Amr Dec 30 '12 at 15:21
Sure, it makes sense to take non-negative $w(n)$. The case $w(1) = 1$ and $w(n) = 0$ for $n>1$ is a provocative counterexample, so I don't necessarily want to exclude it. – orlandpm Dec 30 '12 at 15:25
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The answer surely depends on the sequence $(x_n)$. If, for example $x_n\equiv 1$, then $b_n$ will converge to one for any weighting sequence $(w(n))$. – Eckhard Dec 30 '12 at 15:48
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The crucial condition on $w$ is $\sum_n w(n)=\infty$. If it holds, then any tail of the sequence $w(n)$ dominates the head, and therefore the influence of the initial terms of the sequence $x_n$ becomes negligible in the long run. If it fails, then the opposite happens and you can find a convergent sequence $x_n$ for which the weighted averages have a different limit (but they will still have a limit). – user53153 Dec 30 '12 at 20:01

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