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Problem 21 Medium Difficulty

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

$ \sum_{n = 1}^{\infty} \frac {n}{b^n} (x - a)^n, b > 0 $


$$I=(a-b, a+b)$$


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Video Transcript

If you want to know where we get convergence, we can use the ratio test so limited n goes to infinity of absolute value of and plus one over a n. And then we want for this to be less than one and by end remained this whole chunk here, including the X values. So this is limit as n goes to infinity, absolute value. And this thing is going to simplify tio X minus a in plus one over B to the n plus one. Time's beat the end over in case there's still some more simplifying too, you hear, And then this be the and divided by B to the n plus one. There's gonna be some cancellations happening there, so that's going to turn out to be just one over B and then we stop this n plus one over in as in goes to infinity and plus one over n goes toe one. So this is absolute value of X minus a overby, as we mentioned, we want this to be less than one. So if we multiply both sides by B and we get absolute value of X minus, A is less than be so already, we can probably see that the radius of convergence is B And we know that because X minus a is trapped between minus B and positive B. So if we take this, we can add eight of both sides. And then the length of the interval of convergence is a plus B minus a minus. B. This is to be and one half of the length of the interval of convergence is just B as we were predicting it would be. Remember, the radius of convergence is one half of the length of the interval of convergence. We know that we get convergence for X between a minus B in a plus B and to figure out the interval of convergence, we just need to figure out whether or not these endpoints work. Okay, so B is our radius of convergence. So now we check to see what happens when X is a minus. B When we plug in a minus B in here we get and over be the end and plugging in a minus B the A and the minus A will cancel. So we have minus B to the end minus B to the N is minus one to the end Times be to the ends and I will get a cancellation with the beat of the in terms For now, I'Ll just turn out to be some let's see some from in equals one to infinity of in And as we mentioned, nothing will cancel out with this minus sign So we'Ll still have minus one to the end but the B to the end will cancel And this is definitely going to be divergent because these terms are certainly not going to be going to zero Okay, And then similarly, if X is equal there a plus B, the vaccine tickled a plus B we plug in X equals a plus B here a minus a zero. So we just have to be here. So be the end divided by B to the end Those will cancel on we just get some from n equals one to infinity of end And again, this is certainly not going to go to zero, So we can't possibly have convergence. So get in, get divergence here. So both of the end points we will not include in our interval of convergence are interval of convergence. We have this open interval here open because we have to throw out this point. We have to throw out this point.

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