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# Find the radius of convergence and interval of convergence of the series.$\sum_{n = 1}^{\infty} \frac {b^n}{\ln n} (x - a)^n, b > 0$

## $R=\frac{1}{b} \quad$ interval of convergence: $\left[a-\frac{1}{b}, a+\frac{1}{b}\right)$

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##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

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### Video Transcript

Okay, So if it is equal to one, then natural log of one is zero. So we'd actually be dividing by zero here. So if there's not a typo than the radius of convergence and the interval of convergence don't exist because no matter what you plug in here, you're gonna have trouble. Ah, but assuming that there's ah type of here and this is supposed to start at something like to then we could figure this out by using the ratio test to figure out where we get convergence. So I guess I'll use ah, different variable like sea. So with you, n goes to infinity of absolutely of C N plus one over CNN just because we already have a CZ happening over here. But by seeing we mean this whole thing, including the X values. So this is limit as n goes to infinity? Absolutely. How you of X minus a And then we'd have times being plus one over being which is just simplifies to be. And then we have Ellen of in over ln of in plus one. So from here to here, I skipped a little bit of the algebra, but I don't think that should be a problem. Just remember that dividing by something is the same thing as multiplying by the reciprocal. Okay. And now we need to figure out what happened. As n goes to Infinity of Ellen of n Divided by Elena van plus one, we can use low Patel's rule here, the derivative of the top, divided by the derivative of the bottom. And we're allowed to use low petal because both the top and the bottom are blowing up to infinity. If they were both going to zero, that would be another set up for low. Patel's rule, so derivative of natural log in is one over in derivative of natural light of N plus one is one over in plus one. So this is limit as n goes to infinity of n plus one over in, which is just one. So this is one. So now this whole thing just turned out to be absolute value of X minus a times B. And we're trying to figure out when this is less than one, so we can multiply both sides by one over B, and then we get absolute value of X minus. A is less than one over B So X minus is trapped between minus B and B. We can add a to both sides. Get a minus one over B is less than acts is less than a plus one over b. Okay, so the radius of convergence here we'd have a plus one over B and then we'd be subtract in a minus one over B. That's the length of her interval. The radius of convergence. We'd have to divide it by two. These ays will cancel and then we'LL get one over b plus one over. Be so we'LL get to Overby and were divided by two It's our radius of convergence is just one over b To figure out the interval of convergence, we need to figure out whether or not we include this end point and whether or not we include this in point. So look to see what happens when axes a minus one over b and then we'LL look to see what happens when X is a plus one over b So when X is a minus one over b, we get a minus, a will cancel and then we have minus one over B to the end so that be the end of the beat of the annual cancel. So then we'd end up with this minus one here. Nothing canceled out with the negative sign and we're dividing by natural lot of in. So this will converge by the alternating signed test. That's good. When X is equal to a plus one over B, you plug in a plus one over B here a plus one over B minus A. We just get one over b. So we went over B to the end, will cancel out with this B to the end. So then we just get one over natural log of n and that's going to be bigger than some of just one over end. Because natural law Gauvin is less than in. And if we replace the denominator with something smaller, then we get something bigger by smaller women, an absolute value. Okay. And this is the harmonic Siri's that blows up. Therefore, this blows up a cz well, so this gives us divergence. So we include a minus one over b. Sorry, we include that because that gives us convergence. But then we toss out a plus one over be so toss it out. So leave that open. That's our interval of convergence. And just as a reminder, our radius of convergence was mentioned over here. One over B.

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##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

Lectures

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03:49

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