Suppose I know $a_n \to 0$ as $n \to \infty$. How fast can $\Sigma_{n=1}^{\infty}a_nr^n$ blow up as $r \to 1$?
In particular, can I say $(1-r)\Sigma_{n=1}^{\infty}a_nr^n \to 0$? If so, how can I prove it?
Also, is there a generally best way to look at questions of this type. I'm very frustrated because I feel like I should be able to answer this myself, but I don't know how to approach it.
$\newcommand{\eps}{\varepsilon}$If $(a_{n}) \to 0$, then for every $\eps > 0$, there exists a positive integer $N$ such that $|a_{n}| < \eps$ for $n \geq N$. If $|r| < 1$, then \begin{align*} \left|(1 - r)\sum_{n=0}^{\infty} a_{n} r^{n}\right| &= \left|(1 - r)\sum_{n=0}^{N} a_{n} r^{n} + (1 - r)\sum_{n=N+1}^{\infty} a_{n} r^{n}\right| \\ &\leq \left|(1 - r)\sum_{n=0}^{N} a_{n} r^{n}\right| + \eps\left|(1 - r)\sum_{n=N+1}^{\infty} r^{n}\right| \\ &= \left|(1 - r)\sum_{n=0}^{N} a_{n} r^{n}\right| + \eps|r^{N+1}| \\ &< \left|(1 - r)\sum_{n=0}^{N} a_{n} r^{n}\right| + \eps. \end{align*} Since $\eps > 0$ was arbitrary, the left-hand side approaches $0$ as $r \to 1$.