# convergence of weighted average

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
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
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

Weighted averages belong to the class of matrix summation methods.

Define

$$W:=\left(\begin{matrix}W_{1,1},W_{1,2},\ldots\\W_{2,1},W_{2,2},\ldots\\\vdots\\\end{matrix}\right)$$

Represent the sequence $\{x_n\}$ by the infinite vector $X:=\left(\begin{matrix}x_1\\x_2\\\vdots\end{matrix}\right)$, and $\{b_n\}$ by the vector $B:=\left(\begin{matrix}b_1\\b_2\\\vdots\end{matrix}\right)$. Then we have $$B=WX.$$

In our case $W_{i,j}:=\frac{w(j)}{w(1)+\ldots+w(i)}$, for $j\leq i$, and $W_{i,j}:=0$, for $j>i$.

The summation method is called regular if it transforms convergent sequences into convergent sequences with the same limit. For matrix summation methods we have Silverman-Toeplitz theorem, that says that a matrix summation method is regular if and only if the following are satisfied:

1. $\lim_{i\rightarrow\infty} W_{i,j}=0$, for every $j\in\mathbb{N}$ (entries converge to zero along columns)

2. $\lim_{i\rightarrow\infty}\sum_{j=1}^{\infty}W_{i,j}=1$ (rows add up to $1$)

3. $\sup_{i}\sum_{j=1}^{\infty}|W_{i,j}|<\infty$ (the sums of the absolute values on the rows are bounded.)

In your case $2$ and $3$ are satisfied (if you assume, as in the comment that $w(i)\geq0$), therefore you get the result if and only if

$$\sum w(i)=\infty.$$

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