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Main content
Current time:0:00Total duration:8:26
AP.CALC:
LIM‑8 (EU)
,
LIM‑8.D (LO)
,
LIM‑8.D.1 (EK)
,
LIM‑8.D.2 (EK)
,
LIM‑8.D.3 (EK)
,
LIM‑8.D.4 (EK)

Video transcript

so we have an infinite series here and the goal of this video is to try to figure out the interval of convergence for this series and that's another way of saying for what x-values what what range of x-values is this series going to converge and like always pause this video and see if you can figure it out so when you look at this series it doesn't fit cleanly into something like a geometric series or an alternating series and when I see something like this I think about the ratio test because it tends to be pretty general and to apply the ratio test we want to think about the limit the limit as n approaches infinity of the n plus 1 term divided by the nth term and the absolute value of that and if this thing is less than 1 so when this thing is less than 1 then we are going to converge and the x-values that make this thing greater than 1 we are going to diverge and the x-values that make this equal to 1 well then we're going to be inconclusive and so we're gonna have to use other techniques to think about whether we're going to converge or diverge so let's just think about this let's just evaluate this so let's do that limit as n approaches infinity of the absolute value of a sub n plus 1 well that's going to be X to the n plus 1 and let me color code this just so we know what we're doing so this thing right over here is going to be X to the n plus 1 over n plus 1 times 5 to the n plus 1 and we're going to divide that by the nth term we're going to divide that by the nth term and that's just going to be X to the N over N times 5 to the N and we're going to take the absolute value of this whole thing now let me simplify this I'll simplify it down here so this is the same thing as X to the n plus 1 over n plus 1 time five to the n plus one times the reciprocal of this it's going to be n times 5 to the N over X to the N and we could simplify this this is going to be equal to let's see you divide numerator denominator by X to the N you're left with just an X and then divide numerator and denominator by 5 to the N that is going to be that's a 1 this is a 1 and then this is just going to be 5 to the n plus 1 divided by 5 to the N that's just going to be a 5 and so what do we have we have x times n x times n / distribute the 5 5 n 5 n plus 1 oh let me be careful there let me distribute the 5 5 n plus 5 5 times n 5 times 1 5 n plus 1 5 5 n plus 5 all right so let me just rewrite that this is going to be equal to the limit as n approaches infinity of the absolute value of this thing and actually to help us with this limit let me rewrite a little bit let me divide the numerator and the denominator both by n I'm not changing the value I'm doing the same thing to the numerator in the denominator I'm dividing it by the same value so if I divide the numerator in the denominator by n this is this it's going to be the same thing as x over 5 plus 5 to the N and so when you I divided the numerator denominator by n it becomes very clear what happens as n approaches infinity as n approaches infinity X doesn't change 5 doesn't change but 5 over n goes to 0 and so this limit is going to be equal to x over 5 so that's a pretty neat clean thing so now we can use this to think about it actually let me write this this is going to be the absolute value of x over 5 and so now we can think about under which conditions is the absolute value of x over 5 going to be less than 1 and we're going to definitely converge under what conditions are we going to be greater than 1 and definitely diverge and then under what conditions is it inconclusive so let's just see when we know we can convert so the absolute value of x over five is less than one this is our convergence situation well that's the same thing as saying that X negative 1 is less than X over 5 which is less than 1 and you multiply all the sides by 5 this is the same thing as negative 5 is less than X which is less than 5 so if we know that this is true this is definitely going to be part of the interval of convergence we know that if X meets these constraints then our series is going to converge but we're not done yet we have to think about the inconclusive case so let's think about the scenario where the absolute value of x over 5 absolute value of x over 5 is equal to 1 or another way of thinking about this this means the that means x over 5 is equal to 1 or x over 5 is equal to negative 1 and this means that x is equal to 5 or X is equal to negative 5 so these are the two inconclusive cases using the ratio test so let's test them out individually by looking back at the series and just substituting x equals 5 or x equals negative 5 so in the first scenario let me find a new color here I'm going to use red so in the first scenario of x equals 5 let's go to our series then the series is going to be the sum from N equals 1 to infinity of 5 to the N over N times 5 to the N well this is just going to be equal to the sum N equals 1 to infinity of 1 over N and this is a harmonic series this is the P series where P is equal to 1 and we know if our P series of P is equal to 1 that's going to diverge so then and we know the harmonic series we've done in other videos this definitely diverges so this diverges and you can do that by P series convergence test if the P for P series is 1 well you're going to diverge so now let's think about so 5 is definitely part of its inter part of our interval of convergence now think about x equals negative five when x equals negative five let me get a another color going here when X is equal to negative five then this thing is going to be equal to the sum from N equals one to infinity of negative five to the N and actually let me just write that as I'll write it out negative five to the N over N times 5 to the N this is the same thing as the sum from N equals 1 to infinity of we could write this as negative 1 to the N times 5 to the N times 5 to the N over N times 5 to the N and now this thing this is an alternating harmonic series and so you can actually use the we you might already know that that converges or you could use the hot you could use the alternating series test and the alternating series test it might be a little bit clearer if I write it like this that this is an alternating series so in an alternating series test if we see if we see that this thing is monotonically decreasing and the limit as n approaches infinity is 0 this thing converges the alternating harmonic series actually converges so this converges so given that this converges you could view this as this boundary here we would include that in our interval of convergence so X doesn't just have to be strictly greater than negative 5 it could be greater than or equal to negative 5 but it has to be less than 5 this is our true interval of convergence
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