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Find the radius of convergence and interval of co…

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Problem 12 Medium Difficulty

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

$ \sum_{n = 1}^{\infty} \frac {( - 1)^{n-1}}{n5^n} x^n $

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Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 11

Infinite Sequences and Series

Section 8

Power Series

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Video Transcript

the radius of convergence are is the limit as n goes to infinity of absolute value of a n over a n plus one were the ends are these terms here, So a and is minus one of the n minus one divided by n times five to the end. Once we write this in, we get one over n times five to be in and then dividing by 8 p.m. plus one is the same thing as multiplying by the reciprocal so multiplying by the reciprocal would get times n plus one times five to the end plus one. And keep in mind that this minus one of the n minus one gets thrown out because we're looking at absolute value here. So five to the end, plus one divided by five to the end is just five and limit as n goes to infinity of n plus one over n is one. So this turns into one times five, which is just five. So that's our radius of convergence. For the interval of convergence, we need to ask whether or not we include X equals five and whether or not we include X equals minus five so an X is minus five. The terms that we're going to be summing up are going to be minus one to the N minus one over in times five to the end and then multiplied by minus five to the end. Okay. And those, there's going to be some nice cancellations happening here. So minus one to the N minus one times minus one to the end is just minus one five to the end. Divided by five to the end is one. So this is going to be minus one over n. So where summing this up, this is going to be the minus version of the harmonic series. So that's going to diverge. Sex equals minus five. We will not include in our interval of convergence. And then when X is equal to five, we plug in X equals five in here and we get something similar. Except now we have minus one to the N minus one, divided by and and this is going to converge by the alternating sign test. So minus five, we get divergence. Five, we get convergence. So the interval of convergence we throw out minus five. We include five

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