Find the radius of convergence, R, of the series. and Find the interval, I, of convergence of the series.

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$$ \sum_{n=1}^{\infty} 3^n x^n / n^2.$$

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?

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1 Answer

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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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