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Find the radius of convergence and interval of convergence of the series.

$ \sum_{n = 1}^{\infty} \frac {( - 1)^n x^n}{n^2} $

$R=1 \quad[-1,1]$

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Oregon State University

University of Michigan - Ann Arbor

Idaho State University

Boston College

case, Remember, our radius of convergence are a cz limit as n goes to infinity absolute value of a N over A and plus one where these guys are a in terms. So we get Lim as in goes to infinity the minus one to the end that'Ll just get thrown out when we put these absolute value signs on So we're going to have absolute value of in plus one squared, divided by in squared because dividing by something of the same thing as multiplying by the reciprocal. That's how the plus one squared ended up on top and this limit is one. Now, for the interval of convergence, we need to figure out whether or not one or minus one works for convergence when we plug it into here. We just checked both of those individually when x is minus one, we have minus one of the end times minus one to the end, so that would just give us a positive one. And this is something that converges because the exponents, too, is bigger than once. This that's convergent. So that works and then X equals one. You have minus one to the end over in squared this is certainly going to converge. We've in fact, just shown that it'll converge. Absolutely so certainly will converge. Sex equals one also works. So both X equals minus one and X equals one. Both work. So we include both of those when we construct our interval of convergence. So when we include the points, we have these close brackets here.