# Prove $\sum_{n = 0}^{\infty} a_n X^n$ converges at every point in $[-1,1)$ if $a_n$ is non-increasing and converges to $0$

Prove that if $a_n$ is non-increasing, converges to $0$, and radius of convergence of $\sum_{n = 0}^{\infty} a_n X^n$ is equal to $1$ then this series converges at every point in $[-1,1)$

I am trying to work out this proof using abel's theorem, but I missed the classes where its use as a conditional convergence test was demonstrated. Once I have an idea of how this first proof should go, I think I can also prove a similar result if $a_n$ is non-decreasing on $(-1,1]$ If I am mistaken and using Abel's theorem here is not as easy as Dirichlets or Leibniz then guidance for that direction would also be appreciated.

• Do you mean $\sum a_n X^n$? Otherwise, what is the radius of convergence of a series of real numbers? Commented Apr 25, 2016 at 0:38
• @AloizioMacedo yes, I'm sorry that was silly. Commented Apr 25, 2016 at 0:41
• It looks like the standard result about "alternating" series. Commented Apr 25, 2016 at 0:44
• Yep. You already know that the series converges for all $X$ in $(-1,1)$, by the radius of convergence information; all you have to do is to establish convergence for $X=-1$. Commented Apr 25, 2016 at 0:45
• $a_n$ being positive and non-increasing actually implies that the radius of convergence is (at least) one. Commented Apr 25, 2016 at 0:56

I'm just going to use $x$ instead of $X$. Since the radius of convergence is $1$, we know that the series converges for every $x\in(-1,1)$. So to show that the series also converges for $x=-1$ we can use the alternating series test since the series would look like $\sum_{n=0}^\infty (-1)^na_n$. The alternating series test says that the series $\sum_{n=0}^\infty (-1)^na_n$ will converge if the sequence $a_n$ is not increasing (monotonically decreasing) and $\lim_{n\to\infty} a_n=0$. Since these conditions are met by hypothesis, we see the series will converge for $x=-1$, however we don't know about when $x=1$.