Radius of convergence and the endpoints of a power series

Find the radius of convergence and the convergence at the end points of the series:

$$\sum_{n=1}^\infty(2+(-1)^n)^nx^n$$

This is what I did:

$a_n=(2+(-1)^n)^n\Rightarrow R=\frac{1}{limsup|a_n|^\frac1n}\\ for\ n=2k \to lim(2+1)=3 \\ for\ n= 2k+1 \to lim(2-1)=1 \\ limsup \ a_n^{\frac1n}=3 \Rightarrow R=\frac13$

So the series converge at $(-\frac13 , \frac13)$

But now I don't really understand what I'm being asked for the end points. Am I supposed to check if the series really converge at the two end points ?

• Your computation shows that the series converges in the open interval $(-1/3,1/3)$ and diverges if $|x|\gt 1/3$. It leaves unsettled the question of what happens at $-1/3$ and at $1/3$. You need to tackle these questions, since the Root Test does not settle the matter. – André Nicolas Dec 7 '13 at 17:47
• You mean $\limsup a_n^{1/n}=3$. – Jean-Claude Arbaut Dec 7 '13 at 17:47
• I see Andre. Yes, edited arbautjc. – GinKin Dec 7 '13 at 17:49
• @AndréNicolas am I supposed to place, say 1/3 in the X to check for it ? – GinKin Dec 7 '13 at 18:14
• Exactly. Do separately $x=1/3$ and $x=-1/3$. For each of them, write down the first few terms and you will see what's going on. It will turn out that the terms do not have limit $0$, so the series cannot converge. – André Nicolas Dec 7 '13 at 18:18

Your analysis for the radius of convergence is good. To check the convergence/divergence at $x=\pm\frac13$, ask whether the terms go to $0$ as $n\to\infty$.
I would split it into the sum of two series, one with even $n$'s and one with odd $n$'s. One is geometric with common ratio $x^2$. The other is geometric with common ratio $9x^2$. Both must converge (since the power series are positive for positive $x$), so applying the Ratio test to the sum of the $(9x^2)^{n}$'s gives you a radius of convergence of $1/3$ and a radius of convergence of $1$. for the sum of the $x^{2n-1}$'s. Check whether the series converges for $x = \pm 1/3$ by direct substiution into the series.