I'm trying to solve:
Suppose $\sum_{n=1}^\infty a_n$ converges. Prove that: $$ \lim_{r\to1^-}\sum_{n=1}^\infty r^n a_n = \sum_{n=1}^\infty a_n. $$ Hint: Sum by parts.
In class, I have seen a proof that involves showing first (using summation by parts) that $\sum_{n=1}^\infty r^na_n$ is convergent for $r\in[0,1]$ rather than for $r\in[0,1)$ only, and then exploiting continuity to obtain the result.
My question: is there a more direct way, as the text would suggest, to use summation by parts in order to isolate the $r\to1^-$ limit from $N\to\infty$? I tried both ways of performing partial summation: $$ S_N=\sum_{n=1}^N r^n a_n = r^N \sum_{n=1}^Na_n+ (1-r)\sum_{n=1}^{N-1}r^n\sum_{j=1}^{n}a_j\\ =a_N\frac{1-r^N}{1-r}+\sum_{n=1}^{N-1}(a_n-a_{n+1})\frac{1-r^{n+1}}{1-r}. $$ Taking $N\to \infty$ first (as we have to...) $$ S=\sum_{n=1}^\infty = (1-r) \sum_{n=1}^\infty r^n \sum_{j=1}^na_j\\ =\frac{1}{1-r}\sum_{n=1}^\infty (a_n-a_{n+1})(1-r^{n+1}). $$ However, I appear to be left with nothing new...