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?
 A: 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:


*

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

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

*$\sup_{i}\sum_{j=1}^{\infty}|W_{i,j}|<\infty$ (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 
$$\sum w(i)=\infty.$$
