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Problem 16 Medium Difficulty

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

$ \sum_{n = 1}^{\infty} \frac {( - 1)^n}{(2n - 1)2^n} (x - 1)^n $

Answer

$R=2 \quad$ interval of convergence: $(-1,3]$

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Video Transcript

to figure out the radius of convergence here will use the ratio test. So take limit as n goes to infinity of absolute value of a and plus one over a n where a n is this whole chunk here, including the X values. So this is limit as n goes to infinity of absolute value of X minus one to the end plus one over two times in plus one minus one times two to the n plus one. So this is our A and plus one, except we've tossed out the minus one to the end And then we're dividing that by a m, which is the same thing. I was multiplying by the reciprocal So now we're multiplying by two and minus one times two to the end and we're dividing by X minus one to the end, x minus one to the end, plus one divided by x minus. One of the inn is just X minus one. So this is Lim is n goes to infinity of absolute value of text minus one and then we have two and minus one over two times in plus one minus one times two and over to N plus, once we just rearrange things a little bit. This limit here is one because we have something linear on top with leading coefficient to and something linear on the bottom with leading coefficient to sew as n goes to infinity, this goes toe one. Here we get a bunch of cancellations that happen and he's going to be left with one copy of two down at the bottom. So this turns out to be absolute value of X minus one over two, and we want for this to be less than one. So x minus one over two is less than one. If you multiply both sides by two, we get absolute value of X minus. One is less than two, which means that X minus one is trapped between minus two and positive too. So we have minus two is less than X minus one, and we add one to both sides. Then we get minus. One is less than sacks and if we have X minus, one is less than two and we add one to both sides. Then we get X is less than three. So X converges between minus one and three. And at this point. We're not yet sure whether or not we get convergence up minus one and whether or not we get convergence at three. But we do know that the length of our interval of convergence is three minus minus one. This value minus this value to the radius of convergence, is the length of the interval of convergence divided by two. So we get four divided by two. So too, so too is our radius of convergence. Here, the interval of convergence. We need to figure out whether or not to include minus one whether or not to include three. So if X is minus one, then we have minus two to the end here. So minus to the minus one to the end, times two to the end It's a minus one of the end times minus one of the and we'LL cancel two to the end, divided by two to the end will cancel. So if X is minus one, we had a bunch of cancellations to happen and then we just have someone over two in minus one, and this is going to diverge because we can compare that Teo, just one over two n this denominators smaller than this denominator. So that means that this value has to be bigger than this value. This is one half of the harmonic. Siri's harmonic Siri's we know blows up. So this is divergence here, So we don't want include X equals minus one for X equals two three if you plug in X equals three here than we have to to the end. So we have to do the and cancer now with the two to the end down there. So then we would have minus one to the end, over to N minus one. And this is going to converge by the alternating signed test. Okay, so three works minus one does not work. So we do not include minus one, but we do include three.

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