Does the series admit a limit as $\alpha\to 0$? I have the following problem:
Let $\{x_n\}_{n\geq 1}$ be a bounded sequence of real numbers and let $\sigma_n=\sigma_n(\alpha)$ be a sequence of weights, depending on a real parameter $\alpha>0$, such that $\sigma_n>0$, $$\sum_{n\geq 1} \sigma_n<\infty$$ for all $\alpha>0$ and $\sigma_n\to 1$ as $\alpha\to 0$. To fix ideas, consider
$$ \sigma_n = \frac{1}{1+\alpha n^2},$$
but I would like to know if there are results more in general, for $\sigma_n$ satisfying the properties above. Now, by the hypothesis we know that
$$\sum_{n\geq 1} \sigma_n x_n$$
is absolutely convergent for any $\alpha>0$. What I'm interested in is understanding if the following rescaled series admits limit as $\alpha\to 0$:
$$\lim_{\alpha\to 0} \frac{\displaystyle \sum_{n\geq 1} \sigma_n x_n}{\displaystyle \sum_{n\geq 1} \sigma_n}.$$
My intuition tells me that, since $\sigma_n\to 1$ as $\alpha\to 0$, it will tend to some sort of uniform distribution, i.e.
$$\lim_{\alpha\to 0} \frac{\displaystyle \sum_{n\geq 1} \sigma_n x_n}{\displaystyle \sum_{n\geq 1} \sigma_n} 
= \lim_{N\to\infty} \frac{1}{N} \sum_{n=1}^N x_n,$$
assuming the RHS is well defined. So the question is: is my intuition correct? Or is it correct under suitable assumptions on $\{x_n\}_n$ or $\{\sigma_n(\alpha)\}$?
This is not an exercise, it's something that has come up in a project I'm working on, and usually I don't deal with this type of problems so I don't know how to approach it. Any help is appreciated!
EDIT: when I talk about assumptions on $\sigma_n$ I mean possiblle assumptions in addition to the ones I'm already imposing. Moreover, I'm more interested in understanding if the limit exists and what it looks like, rather then force the assumptions in order to make it become what my intuition tells me it should be. To make it clear: I'm mainly interested in the case
$$\sigma_n=\frac{1}{1+\alpha n^2}$$
but there are lots of other choices of $\sigma_n$ which satisfy the assumptions, as for example
$$\sigma_n=\frac{1}{1+\alpha n^c}\ \text{ for some } c>1$$
but also things like
$$\sigma_n=\frac{1}{1+\alpha n!}$$
with the latter example decaying much faster then the others. So I'm trying to understand what would be the limit of the ratio in the first example, and if it would change taking the other choices or it would be still the same.
 A: If you haven't seen it already, take a quick look at the Wikpiedia article https://en.wikipedia.org/wiki/Divergent_series, and Hardy's book "Divergent series". There is a notational difference between series (as in summability theory) and sequences (as in your question), but it is easy to translate between the two.
If $x_n$ converges to a limit $x,$ then by the (easy direction of the) Silverman-Toeplitz theorem your rescaled limit converges to the same value $x$; use $a_{i,j}=\sigma_j(\alpha_i)/\sum_{n\geq 1}\sigma_n$ for a sequence $\alpha_i\to 0.$ So we're dealing with a "regular" matrix method. Well, sort of - "matrix methods" require $\alpha$ to be discrete.
For $x_n$ not necessarily convergent, if the weights $\sigma_n$ are non-increasing for each $\alpha,$ and the Cesàro mean $x=\lim_{n\to\infty}\tfrac 1N\sum_{n=1}^N x_n$ exists, then the rescaled limit is the same value $x.$ So your rescaled series is "consistent with" the Cesàro mean. To show this, let $m_n=\tfrac 1N\sum_{n=1}^N x_n$ and write $\sum_{n\geq 1}\sigma_nx_n$ as $\sum_{n\geq 1}(\sigma_n-\sigma_{n+1})nm_n.$ The values $(\sigma_n-\sigma_{n+1})n$ are non-negative, and $m_n\to x,$ so by the Silverman-Toeplitz theorem your rescaled limits tend to $x$; use $a_{i,j}=(\sigma_n-\sigma_{n+1})nm_n/\sum_{n\geq 1}\sigma_n.$
