Limit of series where termwise convergence holds Given $a_n\in\mathbb{R}$ such that $\sum_{n=1}^\infty |a_n|^2<\infty$. Is it true that $$\lim_{r\rightarrow 1^-}\sum_{n=1}^\infty(1-r^n)^2|a_n|^2=0?$$
It is certainly true the limit of each term in the series is $0$. But since the series is infinite, is the statement true?
 A: You can directly show what you want without invoking monotone convergence, and that may be what is needed in this case, depending on your background. Let $\epsilon > 0$ be given. One need only show that there exists $\delta > 0$ such that $\sum_{n=1}^{\infty}(1-r^{n})^{2}|a_{n}|^{2} < \epsilon$ whenever $1-\delta < r < 1$. To do this, first notice that, for any positive $N$ and $0 \le r \le 1$,
$$
             0 \le \sum_{n=1}^{\infty}(1-r^{n})^{2}|a_{n}|^{2} \le \sum_{n=1}^{N}(1-r^{n})^{2}|a_{n}|^{2}+\sum_{n=N+1}^{\infty}|a_{n}|^{2}.
$$
Because $\sum_{n=1}^{\infty}|a_{n}|^{2}$ converges, the last sum on the right can be made strictly less than $\epsilon/2$ by choosing some large enough, fixed $N$. The limit of the first sum on the right as $r\uparrow 1$ is 0 for this fixed $N$ and, so, there exists $\delta > 0$ such that the first sum on the right is strictly bounded by $\epsilon/2$ whenever $1-\delta < r < 1$. It then follows that the sum on the left is strictly bounded by $\epsilon$ whenever $1-\delta < r < 1$. Because $\epsilon > 0$ was arbitrary, then, by definition, $\lim_{r\uparrow 1}\sum_{n=1}^{\infty}(1-r^{n})^{2}|a_{n}|^{2}=0$.
A: Monotone Covergence also guarantees the limit you are considering. You simply need to apply Monotone Covergence to
$$
\lim_{r\to1^-}\sum_{n=1}^\infty(2r_n-r_n^2)|a_n|^2
$$
(so the terms increase monotonically) and subtract from
$$
\sum_{n=1}^\infty|a_n|^2
$$
A: Yes, if you have an absolute convergent dominant series, you can exchange limits and summation. This is like the Lebesgue theorem of dominated convergence.
Or look at it another way: each term is a function in $r$. Per assumption, the series of max-norms converges, so you get uniform convergence to the pointwise limit, and the limit function is continuous. Thus the value 0 at r=1 is the given limit.
