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Current time:0:00Total duration:9:37

Video transcript

you might recognize what we have here in yellow as the general form of a p-series and what we're going to do in this video is think about under which conditions under for what Peas will this pea series converge and for it to be a p-series by definition P is going to be greater than zero so I've set up some visualizations to think about how we are going to understand when this pea series converges so over here you have the graph this curve right here that's y is equal to 1 over X to the P and we're saying it in general terms because P is greater than 0 we know it's going to be a decreasing function like this once again that's y is equal to 1 over X to the P and now what we've shaded in ahead of time underneath that curve above the positive x-axis that is the integral from 1 to infinity from 1 to infinity improper integral of 1 over X to the P DX so that's this area that I have already shaded in you see it in white in both of these graphs and we're going to hopefully see visually is that there's a very close convergence or divergence relationship between this P Series and this integral right over here because when we when we when we look at this left-hand graph we see that this P series can be viewed as an upper riemann approximation of that area what is that what do I mean by that well think about think about the area of this first I guess you could say this first rectangle the width is 1 and its height is 1 over 1 over 1 to the P so this would be the first term in this piece here is it this would this would just be an area of 1 this the scale the x and y scales are not the same this one right over here it's area would be 1 over 2 to the P this area is 1 over 3 to P so the sum of the areas of these rectangles that is what this P Series is and you can see that each of these rectangles they are covering more than the area under the curve and so we know the area under the curve that's going to be greater than 0 this P series is going to be greater than this integral greater than the area under the curve but if we add one to the area under the curve so now we're not just talking about the wide area we're also talking about this red area here well then our piece Erie is going to be less than that because the first term of our P series is equal to one and then all of the other terms you can view it as a lower riemann approximation of the curve and you can see they fit under the curve and they leave some area so this is going to be less than that expression there now think about what happens if we know if we know that this right over here diverges so if this improper integral diverges it doesn't converge to a finite value well this the P series is greater than that so if this diverges then that's going to diverge similarly if this converges the same integral right over here if this converges it goes to a finite value well 1 plus that is still going to converge and so this the our P series must also converge it must go to a finite value and what I'm all on one all I'm talking about right here this is really just the integral test when we when we think about tests of convergence and divergence but I'm just making sure that we have a nice conceptual understanding and not just blindly applying the integral test so and you could go the other way too if the P series converges then for sure this integral is going to converge and if the P series diverges and for sure this expression right over here is going to diverge on D and the integral diverges so we can say the P series converges if and only if this integral right over here converges so figuring out under what conditions for what P does the P series converge where it's boiling down to under what conditions does this integral converge so let's scroll on down to give us some real estate to think about what has to be true for that integral to converge so I'm going to rewrite it so we've got the integral from 1 to infinity improper integral of 1 over X to the P DX this is the same thing this is the limit as I'll use the variable M since we're already using n as m approaches infinity and the integral from 1 to M of one over and actually let me just write that as X to the negative P X to the negative P DX and let me just focus on this and we'll just remember that we're gonna have to take the limit as M approaches infinity I don't have to keep writing that over and over again so let's think about what this is so there's a couple of conditions we know we already know that P is greater than zero we know that P is greater than zero but there's two situations right over here there's one situation when P is equal to 1 if P is equal to 1 then this is just the integral of 1 over X and so this thing is going to be the integral of Ln of X and we're going to go from 1 to M and so this would be the natural log of M minus the natural log of 1 well e to the 0 power is 1 so the natural log I'll write it out the natural log of 1 but the natural log of 1 is just 0 so when in the in a special case I guess you can say when P equals 1 this integral from 1 to M comes down to the natural log of M now let's think about the situation where P does not equal the where P does not equal 1 well there were kind of just reversing the power rule that we learned in basic differentiation so we increment that exponent so it would be X to the negative P plus 1 and we could even write that as X to the 1 minus P that's the same thing as negative P plus 1 and then we would divide by that so 1 minus P and we would go we are going to good we are going to go from 1 to M and so this is going to be equal to we could write this as m to the 1 minus P over 1 minus P minus 1 to the 1 minus P over 1 minus P so now let's take the limits so we remember this integral we won't take the antiderivative or the definite integral here but then we want to the limit as n approaches infinity so what is the limit as M approaches infinity of natural log of natural log of M well if M go unbounded to infinity well the natural log of that is still going to go to is still going to go to infinity so when P equals 1 this thing doesn't converge this thing is just unbounded so P equals 1 we diverge so we know that so now let's look over here let's think about the limit as M approaches infinity of this expression right over here and the only part that's really affected by the limit is the part that has M so we could even write this as we could take this 1 over 1 minus P out of this we could say 1 over 1 minus P times the limit as n approaches infinity of m to the 1 minus P and then separately we can subtract 1 to the 1 minus PE well for any exponent that's just going to be 1 over 1 minus P is that right yeah no matter what exponent I put up here 1 to any power is what's going to be 1 and so the interesting thing about whether it converges or not is this part of the expression right over here it's all going to depend on whether this exponent is positive or negative if 1 minus P is greater than 0 well if I'm taking a a if I'm going to infinity and I'm taking that thing to a positive exponent well then this is going to diverge and so in this situation we diverge and 1 minus P is greater than 0 we can add P to both sides that's the situation that's the same thing as 1 being greater than P or P being less than 1 we are going to diverge so far we know that P is going to be greater than 0 and so we saw if P is 1 or if it's less than 1 we're going to diverge but if this exponent right over here is negative if 1 minus P is less than 0 well think about it then it's going to be then it's going to be 1 over m to some positive exponent is one way to think about it so as m approaches infinity this whole thing is going to approach 0 so this is actually going to be a situation where we converge where we get to a finite value and so we add P to both sides we have 1 is less than P we converge so there you have it we have established this integral is going to converge only in the situation where P is greater than one P greater than one you are going to converge and if zero is less than P is less than or equal to one you are going to diverge and those are then the exact because this the R P series converges if and only if this integral converges and so these exact same constraints apply to our original P series our original P series converges only in the situation where P is greater than 1 then we converge and if zero is less than P is less than or equal to 1 we diverge there you go
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