convergence of weighted average
Solution 1:
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
$$\sum w(i)=\infty.$$