 Hello and welcome to a screencast today about finding the integral of convergence for a power series. Okay, so the power series we're going to look at today is the sum as k goes from 1 to infinity of x minus 1 to the k divided by k plus 2 times 4 to the k, and that 4 to the k is in the denominator, just so you know. Okay, on Geogebra I graphed our different partial sums for this particular power series. k equals 15 is in red, so that's kind of this outer function down here. k equals 25 here is in blue, kind of has the basic same shape to it, but maybe a little bit tighter. And then 50 looks totally different, well kind of totally different. On one end it looks different. It's here in black. So you can kind of see then, so our power series is centered around 1 because we're subtracting a 1 off of that. So then it kind of looks like things start to go apart here at negative 3, and sorry you can't really read that number, but that's negative 3, and then maybe somewhere in here in the 4 to 5 range, things kind of start to shoot off too. So it gives us a good visualization of what's happening. So we can kind of have a gut check maybe to see if our interval of convergence is correct or not. Okay, so here's our power series, and again remember we want to go ahead and use our ratio test. So let me remind us of what that is. So we want to look at the ratio test, which says we've got the limit as k goes to infinity of the absolute value of a sub k plus 1 all over the absolute value of a sub k. And remember this is going to converge when this limit is less than 1, which is obviously what we want. We want the interval of convergence, so that's really what we're looking for here. Okay, so let's go back, oops I guess I shouldn't have gone down too far here. Here is our a sub k, all the stuff here inside our summation. So a sub k plus 1 just means that we're going to add 1 wherever we see a k. So that's going to give me, and then if you've watched a video before with these absolute values, I'm going to go ahead and take some liberties here with them. So I'm going to write the absolute value of x minus 1 to the k. K is always positive, so the absolute value of that's not going to affect anything. Divided by, same thing with our denominator, okay, oops and I forgot to add my 1. Got too busy talking about the absolute values. In our denominator the same thing is going to happen, right? Everything here is positive, so those absolute values are not going to affect anything. So I'm going to go ahead and write k plus, let's see, so k plus 1 plus 2 is going to give me k plus 3, and that's a k, and then we're going to have 4 to the k plus 1, okay? Alright, that's just in our numerator. Then we're going to divide by, I'm going to do something kind of similar in my denominator, so that's going to look like the absolute value of x minus 1 to the k, and then divided by k plus 2 times 4 to the k. So again, I'm only putting the absolute values on the pieces that it might affect, okay? Alright, well this is just a big fat mess, so I see a fraction over a fraction, so let's go ahead and multiply by the reciprocal of all of this stuff down here. Okay, so I'm going to have the absolute value x minus 1 to the k plus 1, all over k plus 3 to the 4k plus 1, and then I'm going to multiply, flip this guy over, so I'm going to have k plus 2, 4 to the k, all over the absolute value of x minus 1 to the k. Alright, well this looks like a big mess too, which it kind of is. So let's go ahead and break it apart and kind of group like things together. So that's, we still got our limit, k goes to infinity, I don't want to lose that. So I've got my absolute value of x minus 1 to the k plus 1, and then oops, I lost my absolute value. Then in my denominator, I've got the absolute value of x minus 1 to the k, right? Those two pieces are like, and we're multiplying all this junk together, so it doesn't matter how we break it apart. Okay, I've got times, let's go for that k plus 2, and then I see a k plus 3 in my denominator, that's kind of similar. And then times, I've got my 4 to the k in the numerator, and then my 4 to the k plus 1 in the denominator. Alright, now we're getting somewhere. This is looking much nicer than I did before. Okay, so we've got the limit, k goes to infinity, and now we've got to use some properties of exponents here. So hopefully you remember that. And in this first piece, this x minus 1 part, we're going to be subtracting those exponents because we're dividing, so that's just going to leave us with an x minus 1 in absolute value. The second piece here, I can't really do much with that, so let's just rewrite that. And then the third piece here, we can definitely use our properties of exponents again, which is going to leave us with 1 fourth. Woohoo, alright, that's nice. Okay, so I'm going to box this area off over here, sorry. So we've got the 1 fourth and the x minus 1 no longer have anything to do with k's, so let's go ahead and take those out front. And then we've got our limit, k goes to infinity, k plus 2 over k plus 3. Okay, so now using L'Hopital's or dividing each piece by k, however you want to look at it, hopefully you see that this limit just goes to 1, right? That's not so bad. So we've got the 1 fourth x minus 1 left, and remember I said this converges when all this stuff is less than 1. So that means that the absolute value of x minus 1 then must be less than 4. Okay, so make ourselves a number line. Stick 1 in here. We're going to be going 4 to the right, which leaves us at negative 3. Not a big surprise, that's what we saw in our graph. And if we go 4 to the right, that's going to take us to 5. And again, not too big of a surprise. Remember I said it looked like 4 between 4 and 5, something funky was going on? Okay, so now this is where we actually have to test these two values, and we need to see does our series converge or diverge at those specific values? Okay, all right, let me go down here. So we've got x equals negative 3. I'm going to plug that into our series, so that gives me the series k goes from 1 to infinity of, let's see, so remember it was x minus 1 to the k, so negative 3 minus 1 is going to give us a negative 4 to the k, all over k plus 2, 4 to the k. All right, again, let's do a little bit of algebra here. So hopefully you agree that I can break apart this negative 4 to the k and I can write it as negative 1 to the k times 4 to the k because I want to cancel this stuff in my denominator. Okay, so 4 to k's cancel. I end up with negative 1 to the k over k plus 2. This is an alternating series, obviously that's going to go to 0, so this thing converges. All right, so negative 3 were good, and again, looking at our graph, that made sense, right? At negative 3, it looked like they were all kind of clustered in there together. Okay, x equals 5. Oopsie, I don't know what happened there, let me try that again. So I've got my sum, k goes from 1 to infinity. I'm going to plug 5 in, so 5 minus 1 gives me 4 to the k, all over k plus 2, 4 to the k. Again, my 4 to the k's are going to cancel, so that's going to leave me with, oh jeez, my sigmas are getting worse and worse. k goes from 1 to infinity of 1 over k plus 2. Does this converge? Does this diverge? I don't know. It kind of looks like the series of 1 to the k, so let's use that limit comparison test. This is a good time to refresh some of these tests. There are lots of other tests you can use on this one, but this is the one I'm choosing to use. So I want to compare it with the series, k goes from 1 to infinity of 1 over k. That's a harmonic series, and that one's going to diverge, just for the record. So the limit comparison test says everything has to be positive. Check. Now we need to do the limit of, we're going to divide these guys out. So we want the limit, k goes to infinity of 1 over k plus 2 over 1 over k. Again, we got fractions over fractions, so we want to multiply by our reciprocal. I'm sure you guys can see this is a lot of algebra with this stuff. Okay, take the limit as k goes to infinity. Again, we've got k's in the numerator and the denominator, so this limit's going to go to 1. 1 is a positive finite number. Okay, so that's a good thing. So that means whatever the series we were comparing it to did, our original series does too. So this series diverges, so that means this series must diverge as well. Okay, which again makes sense because it looks like at 5 stuff shooting up to infinity. Okay, so our interval of convergence then is going to be negative 3, and I want to include that because that one did converge to 5, but I'm not going to include that, so I'm going to put a parenthesis there because that one diverged. Alright, so there's lots of pieces here with these, and you've got to put all those pieces together of stuff that we've done before. Alright, thank you for watching.