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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}{\sqrt[3]{n}} $

$$[-1,1)$$

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okay for this problem. To get the radius of convergence, we take the limit as and goes to infinity, would you? Absolute value of hey n over and plus one. These guys here are a in terms, so we're going to get limit as n goes to infinity of absolute value of Q brute of n plus one over q brute of n So this is limit as n goes to infinity of the absolute value of Q brute of n plus one over in lemon is n goes to infinity of implicit one over in his one. So this is Cuban of one. So this is just one. So for the interval of convergence, we need to figure out whether or not we include one and whether or not we include minus one. So figure out what happens when X is equal to minus one and figure out what happened to next sequel, the positive one in here. So if access minus one, we have minus one of the end times a minus one of the ends that just gives us one up top. So then we would just have one over Hugh Brute of N. This is like into the one third power. One third is less than one. Therefore, this diverges. Okay, so we're not going to include minus one here. When X is equal to one, we get some from n equals one to infinity of minus one to the end over Cuba. It of n And this is going to converge by the alternating signed test. So we will include one. We're not gonna include minus one. So are you. Interval of Convergence is going to look like minus one comma one and one. We include minus one. We don't include