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Find the radius of convergence and interval of co…

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Problem 18 Medium Difficulty

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

$ \sum_{n = 1}^{\infty} \frac {\sqrt{n}}{8^n} (x + 6)^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

for the radius of convergence. We use the ratio test to figure out which values of X are allowed the ratio test Would you limit as n goes to infinity of absolute value of a N plus one over an where Anne is this whole thing, including the X values. Okay, so once we plug these things and we get some cancellations to occur, what have ah X plus six to the end, plus one on top. Then we'd be dividing by explosives to the end. So we would just end up with the X plus six and then we'd have an eight to the end of top square root of n plus one over hate to the end, plus one square root of N. Okay, so this is just from doing simple algebra here. Remember our way in terms of this whole thing, including the X values were dividing by a n same thing as multiplying by the reciprocal. So they're super cool of this thing. We're gonna have an Aidan on top as we do, and we're going to have a squared off in the bottom as we do over here. A to the n, divided by a to the n plus one that's just going to turn into one AIDS and then squared of in plus one, divided by squared of end is square root of N plus one over N, then goes to infinity and plus one over N is one squared of one is one. So this is just absolute value of X plus six over eight. And we want for this to be less than one, which is the same thing as saying that absolute value of X plus six is less than eight. Said this point. You could say that the radius of convergence this is eight. And if you really want to be sure about it, um, if we have that absolute value of X plus six is less than eight. That's the same thing as saying that X Plus six is trapped between minus eight and eight, and if we have minus eight, is less than explosives and we subtract six from both sides. We get minus fourteen. If we subtract six from eight, then we get to So the length of the interval of convergence is to minus minus fourteen, which is sixteen. Radius of convergence is half of the length of the interval of convergence. So indeed, we get eight as the radius of convergence The interval of convergence. We just need to figure out whether or not we include these end points here. So when X is minus fourteen, we get convergence or divergence for next of minus fourteen, we're going to end up getting minus eight to the end here minus eight to the end of the same thing as minus one to the end times eight to the end. It will get eight to the end of cancel out with eight to the end and then we'LL have some of the squared of anything here the eight to the angel cancel and we'd have ah, this minus one to the end happening here. But these terms are not even going to go to zero. So we can't possibly have convergence squared of end blows up to infinity. So this, for sure, is not going to converge. We get divergence there. So now we need to check the other end point when X is equal to two, see what happens there when X is equal to two through plug in to here we get eight to the end a to the N divided by eight to the end will cancel out So this can't squared of end Being summed up and again This certainly does not go to zero So we can't possibly have convergence We'LL get the emergence there So both of the end points we have Tio toss out So this is going to be our interval of convergence

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