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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.

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    You've written a general series that could represent many functions. If the function is holomorphic, it *doesn't* blow up at $r=1$ (like $f(r)=e^r$). Other functions might have a pole there or an essential singularity. Did you have any particular function in mind? – 2017-01-16
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    No, I want to know in general. For example, how quickly does it blow up if $a_n=\frac{1}{n}$? What about $a_n=\frac{1}{log(n)}$? – 2017-01-16

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$\newcommand{\eps}{\varepsilon}$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 $r \to 1$.

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    Can you briefly explain your thought process? How did you know how to proceed? Did you ask yourself certain questions? – 2017-01-16
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    Yes: Every convergent (real/complex) sequence is bounded, say $|a_{n}| \leq M$ for all $n$; this gives a geometric series bound (line three, the fundamental idiom of power series). Unfortunately, that's not enough control for the desired purposes. Happily, we can get better control by discarding finitely many terms; in fact, can get an _arbitrarily small upper bound_ by discarding _sufficiently many terms_. Those finitely many terms add up to a polynomial, and the rest are sufficiently controlled. – 2017-01-16