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Determine the radius of convergence of the series solution found in exercise 12 ....

Question

Determine the radius of convergence of the series solution found in exercise 12 .

Determine the radius of convergence of the series solution found in exercise 12 .



Answers

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

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

Okay for this problem. Our radius of convergence are we have limited as n goes to infinity, absolute value of a N over and plus one. And this time that's going to end up looking like in plus one factorial divided by in factorial. Remember the factorial function, in fact or religious like one times two times, three times four times Thought that thought Although I have two times in, there's going to be lots of cancellations happening here. Everything will cancel except for in plus one up top. Okay, to the N plus one will stay there and this limit is clearly infinity. It's our radius of convergence is infinity and typically with the interval of convergence, you would wantto check to see if you know, plus or minus our worked. But in this case, it's ours, just infinity. So our interval of convergence is still just going to be everything all the way. Everything between minus infinity and infinity

Okay for this problem. Our radius of convergence are is lim a as n goes to infinity into the end over in plus one to the end plus one. Okay, so we can rewrite that as limited n goes to infinity of in over in plus one to the power of end and then times one over in plus one. Okay, so in divided by n plus one to the power of n that's one over e and then we're left with limit as in approaches infinity of one over in plus one and this is zero. So we have won over E which is just some finite number time zero So we end up with zero. So the radius of convergence here is zero. The interval of convergence. We just checked the end points on this case. We're just checking zero here and we're checking to see whether or not this is going to converge. There's just adding up zero a bunch of times. So this this is going to be zero to that is going to work. Case will include zero, but nothing else will be included in our interval of convergence Are interval of convergence is just zero all by itself

Hey, to figure out the radius of convergence, we do limit as n goes to infinity of absolute value of a N over absolute value of and plus ones, absolutely of this whole thing where the ends air just one over into the fourth times for the end. This is limit as n goes to infinity Ah, in plus one to the fourth times four to the end, plus one divided by into the fourth times for it to the end. So this's limit as n goes to infinity of thin plus one over in to the fourth and then for the n plus one divided by four to the end Oops, sorry. I forget that for the M plus one divided by four the end that's going to cancel out just before and then and plus one over in as n goes to infinity and plus one over n goes toe wants This is one to the fourth, which is just one. So one times for so just get for here. So four is the radius of convergence Here for the interval of convergence, we need to check whether or not four works when we plug it in here and whether or not minus four works when we plug it in here. So when axes equal to minus four, then we have minus four to the end over into the fourth times for the end. Sorry for the bad hand writing. Okay, so minus four to the end, divided by four to the end, that's going to turn into a minus one to the end because minus forty and is minus one of the end times for the end and then four to the and will cancel out before the end that we have down there. Okay? And then this will converge, but the alternating signed test. And if we look at, if we look to see what happens when X is equal to four, then we would end up with for the end divided by four to the end. So those would cancel. And we just have one over and to the fourth. Since four is a number that is larger than one, this would be something convergent. So this would also converge. Okay, that works. So minus one works are starting minus four works. When you plug it in for works, when we plug it in. So We include both minus four and foreign R Interval of convergence. So we use these close brackets. Tau show that we are including them and that's gonna be our answer.

Since we haven't X to the two and here instead of just an axe to the end, we needto start from the top when we're figuring out the Raiders of Convergence. So as long as you remember what the radius of convergence means and we should be fine and just use the ratio test to proceed the ratio test we look at limit as n goes to infinity of a and plus one divided by a n Now, when we say and we mean this whole thing, including the X values on previous problems, we just meant you know, this whole thing except for the X values that here a n is X to the two and divided by in factorial. Okay, so this turns into limit as n goes to infinity of absolute value of X to the two times in plus one divided by in plus one factorial That's a and plus one. And we're dividing by a m. So we're multiplying by the reciprocal. So we're multiplying by in factorial divided by X to the two n So in factorial is one times two times three times dot, dot dot times in. So if you divide that by N Plus one factorial There's going to be a lot of cancelling that will occur and we'll just be left with in Plus one and then we have X to the two and plus one divided by extra to end so X to the two times in plus one is X to the two n plus two so excellent to and we'LL cancel out with this x to the two n and we'LL just have X squared here. And when you're doing the ratio test you want for this to be less than one. So figure out which values of X is going to make this less than one. Well, all of the values of acts are going to make this less than one. Doesn't matter if X is five ten a million, all all the values of actually going to make this less than one. So our radius of convergence is actually infinity and the interval of convergence is from minus infinity to infinity. Now, if it didn't happen to work out this way, the way you find the radius of convergence, if you only know the interval of convergence is toh, take the length of that interval and then divide it by two. So if this was from minus two, positive to than the radius of convergence would just be, too. In this case, it happens to be infinite and this ratio test here. Remember that in the case where you get equal toe one, you need to be careful when it's equal to one. That test is inconclusive, and you just need to figure out whether or not you get convergence or divergence another way either limit comparison test or a regular comparison test. But when it's equal the one, the test is not conclusive. But this is, Ah, good idea for just figuring out the radius of convergence. And in this case, the radius of convergence is infinity and the interval of convergences minus infinity to infinity. So we get convergence everywhere here.


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