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Current time:0:00Total duration:6:53

AP.CALC:

LIM‑7 (EU)

, LIM‑7.A (LO)

, LIM‑7.A.9 (EK)

let's remind ourselves give ourselves a review of the comparison test see where it can be useful and maybe see where it might not be so useful but luckily we'll also see the limit comparison test which can be applicable in a broader category of situations so we've already seen this we want to prove that the infinite series from N equals 1 to infinity of 1 over 2 to the n plus 1 converges how can we do that well each of these terms are greater than or equal to 0 and we can construct another series where each of the corresponding terms are greater than each of these corresponding terms and that other series the one that jumps out at that will likely to jump out at most folks would be one over two to the N one over two to the N is greater than is greater than and then all we'd have to really say is greater than or equal to but we can actually explicitly say it's well I'll just write it's greater than or equal to 1 over 2 to the n plus 1 for n is equal to 1/2 all the way or all the way to infinity why because this denominator is always going to be greater by one if your denominator is greater the overall expression is going to be less and because of that because each of these terms are they're all positive this one each corresponding term is greater than that one and since and by the comparison test because this one converges this kind of provides an upper bound because this series we already know converges we can say so because this one converges we can say that this one converges now let's see let's see if we can apply a similar logic to a slightly different series let's say we have the series the sum from N equals 1 to infinity of 1 over 2 to the N minus 1 in this situation can we do just the straight-up comparison test well no because you cannot say that 1 over 2 to the N is greater than or equal to 1 over 2 to the N minus 1 here the denominator is lower means expression is greater which means that this can't each of these terms can't provide an upper bound on the when this one is a little bit larger the other hand you're like okay I get that but look as n gets large the two to the N it's going to it's going to really dominate the minus one or the plus one or the or but this one has nothing they're just as two to the N the two to the N is really going to describe the behavior what this thing does and I would agree with you but we just haven't proven that it actually works and that's where the limit comparison test comes in helpful so let me write that down limit limit comparison test limit comparison test and I'll write it down a little bit formally but then we'll apply it to this infinite series right over here so limit comparison test tells us that if I have two infinite series so so this is going from N equals K to infinity of a sub n I'm not going to prove it here we'll just learn to apply it first and this is goes from N equals K to infinity of B sub N and we know that each of the terms a sub n are greater than or equal to zero and we know each of the terms B sub n are actually we're just going to say greater than zero actually can show up in the denominator of an expression so we don't want it to be equal to zero for all the ends that we care about so for all n equal to K k plus 1 k plus 2 on and on and on and on and and this is the key this is where the limit of the limit comparison test comes into play and if the limit the limit as n approaches infinity of a sub n over B sub n B sub n is positive and finite is positive and finite then either both series converge or both series diverge so let me write that so then that tells us that either either both converge or both diverge which is really really useful it's kind of a more formal way of saying that hey look if and as n approaches infinity if these have similar behaviors and they're either going to converge or they're both going to diverge let's apply that right over here well if we say that our B sub n is 1 over 2 to the N just like we did up there 1 over 2 to the N so we're going to compare so these two series right over here notice it satisfies all of these constraints so let's take the limit the limit as n approaches infinity of a sub n over V sub n so it's going to be 1 over 2 to the N minus 1 over over 1 over 2 to the N and what's that going to be equal to well that's going to be equal to the limit as n approaches infinity of twos if you divide by 1 over 2 to the N that's going to it's going to be the same thing as multiplying by 2 to the N it's going to be 2 to the N over over 2 to the n minus 1 over 2 to the n minus 1 and this clearly what's happening in the numerator and the denominator these are approaching the same quantity and actually we can even write it like we can even write it like this divide the numerator and the denominator by 2 to the N if you want well it's probably going to jump out at you at this point so limit as n approaches infinity let me scroll over to the right a little bit if I divide the numerator by 2 to the N I'm just going to have 1 divided by the denominator by 2 to the N I'm going to have 1 minus 2 I can just write this 1 over 2 to the N and now it becomes clear this thing right over here is just going to go to 0 and you're going to have 1 over 1 the important thing is that this limit is positive and finite because this thing is so this thing right over here is positive and finite the limit is 1 is positive and finite if this thing converges in this thing converges if this thing diverges and this thing diverges well we already know this thing converges just so it's a geometric series where the common ratio is less than one and so therefore this must converge as well so that converges as well

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