Showing convergence of a series given convergence of a sequence I am working on a problem which asks me to show the following: Given a sequence of real numbers, $(x_n), n=0,1,2,...$ such that $x_n \rightarrow x$, show that $$\lim_{p\to 1^{-}} (1-p)\sum_{n=0}^{\infty}x_n p^n = x$$
My approach is to try and prove this in a similar fashion to how we prove the geometric series formula (which would be simple if $(x_n)$ were a constant sequence). So looking at the partial sums of the series above, we see that:
$$(1-p)\sum_{n=0}^{N}x_n p^n = x_0 + p(x_1-x_0) + p^2(x_2-x_1) +...+p^N(x_N-x_{N-1})+p^{N+1}X_{N}$$
From here I can't quite let $p\rightarrow 1^{-}$ yet, otherwise everything would cancel. So I want to use the fact that $x_n$ converges to $x$, and I suspect that I'll have to use the fact that since $x_n \rightarrow x$, the $(x_m - x_{m-1})$ terms are going to $0$ for large $m$. However, I still don't know how to deal with the initial terms in the sum where the $(x_m - x_{m-1})$ terms are not negligible.
 A: $\epsilon>0$:
we want to show that there exists a $\delta$ for which if $p\in\left(1-\delta,1\right)$ then $(1-p)\sum_{n=0}^{\infty}x_{n}p^{n}\in\left(x-\epsilon,x+\epsilon\right)$.
we know that x_n converges to x, so there exists an N such that for all n>N we have:
$x_n\in\left(x-\dfrac{\epsilon}{2},x+\dfrac{\epsilon}{2}\right)$. we also know that:
$(1-p)\sum_{n=0}^{\infty}x_{n}p^{n}=(1-p)\sum_{n=0}^{N}x_{n}p^{n}+(1-p)\sum_{n=N}^{\inf}x_{n}p^{n}$. lets look at the second part:
$(1-p)\sum_{n=N}^{\inf}x_{n}p^{n}\geq(1-p)\sum_{n=N}^{\inf}\left(x-\dfrac{\epsilon}{2}\right)p^{n}=\left(1-p\right)\left(x-\dfrac{\epsilon}{2}\right)\dfrac{p^{N}}{1-p}=\left(x-\dfrac{\epsilon}{2}\right)\cdot P^{N}$
so we have:
$(1-p)\sum_{n=0}^{\infty}x_{n}p^{n}\geq(1-p)\sum_{n=0}^{N}x_{n}p^{n}+\left(x-\dfrac{\epsilon}{2}\right)\cdot p^{N}$
but for p that is close enough to 1, the first part goes to zero and the second part goes to x minus epsilon. So you can show for the right delta the lower bound that you need. The upper bound can be shown in a very similar way.
I hope this is understandable
