TITLE: Find the radius of convergence, R, of the series. and Find the interval, I, of convergence of the series.
QUESTION [2 upvotes]: $$    \sum_{n=1}^{\infty} 3^n x^n / n^2.$$
I got $R=1/3$
Find the interval, I, of convergence of the series. (Enter your answer using interval notation.)
I got $I= (-1/3, 1/3)$
i'm fairly certain i got the radius correct, but am not sure about the interval. Could i get some help?

REPLY [2 votes]: It is correct, since you need $(3x) < 1$ which implies $R=1/3$ and center is 0.
At the end points, if $x=1/3$ we have
$$
\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}
$$
and similarly at $x=-1/3$,
$$
\sum_{n=1}^\infty \frac{(-1)^n}{n^2}
$$
converges by the Alternating series test.
The final interval is $[-1/3,1/3]$.
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