# Convergence of a Power series

Consider the power series $\sum^{\infty}_{n=0} a_nx^n$. It is fairly easy to impose conditions on the value of $x$, so as to make the series convergent. However, I was wondering if it is possible to impose conditions on the sequence $\{a_n\}$, so as to ensure convergence of the series, for all $x \in \mathbb R$. Is this possible?

Sure. If $|a_n| \leq \frac{c}{n!}$, where $c$ is a fixed constant, then the series will always converge. The series expansions for $e^x, \sin x$ and $\cos x$ are good examples.
• Thank you for the quick reply. There's a related question I'd like to ask as well:- I understand that this is a sufficient condition for convergence, but do you think this is a necessary condition? Can there be other power series that don't require this condition, yet converge for all $x \in \mathbb R$? – Train Heartnet May 1 '14 at 9:06
• Any power series for which $\frac{|a_{n + 1}|}{|a_{n}|} \to 0$ will converge for all $x \in \mathbb{R}$ by the ratio test. There are certainly other ways to accomplish this without the factorial function. For example, $a_n = \frac{1}{n^n}$. – FlagCapper May 1 '14 at 18:25
• Yes, I understand that. But my question is a bit different; sure, such a condition works for any power series you take. So it's a sufficient condition. But can't there be other power series that converge for all $x \in \mathbb R$ without requiring that condition? I can't think of any examples though. – Train Heartnet May 2 '14 at 4:02
• The ratio test converging to 0 is necessary and sufficient. If the ratio test converges, but not to 0, then the power series will diverge for large enough $x$, because you can choose large enough $x$ to make the limit greater than $1$. To see why this means the series will diverge, you can look at any of the proofs of the ratio test; it basically just relates to the fact that you can always bound the tail of a series by a convenient geometric series. – FlagCapper May 2 '14 at 4:15