Video transcript
- Let's remind ourselves,give ourselves a review of the Comparison Test,see where it can be useful, and maybe see where it mightnot be so useful, but luckily we'll also see the Limit Comparison Test, which can be applicable in abroader category of situations. So we've already seenthis, we want to prove that the infinite series fromn equals one to infinity of one over two to the n plus oneconverges how can we do that? Well, each of these terms aregreater than or equal to zero, and we can construct another series where each of the correspondingterms are greater than each of these corresponding terms. And that other series,the one that jumps out at, that will likely jump outfor most folks would be one over two to the n,one over two to the n is greater than, is greaterthan, and all we would have to really say isgreater than or equal to, but we could actually explicitly say it's, oh I'll just write it'sgreater than or equal to, one over two to the n plus one for, n is equal to one, two, allthe way to infinity - why? Because this denominatoris always going to be greater by one if you'redenominator is greater, the overall expressionis going to be less, and because of that, becauseeach of these terms are, they're all positive, thisone, each corresponding term is greater than that one, andby the Comparison Test, because this one converges,this kinda provides an upper bound, because this serieswe already know converges, we can say, so because this one converges, we can say that this one converges. Now, let's see if we canapply a similar logic to a slightly different series. Let's say we have the series the sum from n equals one to infinity of one over two to the n minus one. In this situation can we do just the straight up Comparison Test? Well no, because you cannot say that one over two to the n isgreater than or equal to one over two to the n minus one. Here, the denominator is lower, means the expression is greater,means that this can't, each of these terms can't providean upper bound on this one. This one is a little bit larger, but on the other hand you'relike, 'Ok, I get that.' But look, as n gets large,the two to the n is going to really dominate the minusone or the plus one, or this one has nothing there,this one just has two to the n. The two to the n isreally going to describe the behavior of 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 LimitComparison 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 what Limit Comparison Test tells us, that if I have two infinite series, so this is going from n equalsk to infinity, of a sub n, I'm not going to prove it here, we'll just learn to apply it first. This one goes from n equalsk to infinity, of b sub n. And we know that each ofthe terms, a sub n, are greater than or equal to zero, and we know each ofthe terms, b sub n, are actually we're just goingto say, greater than zero, because it's actuallygoing to show up in the denominator of an expression, so we don't want it to be equal to zero, for all the n's that we care about. So for all n equal to k, k plus one, k plus two, on and on, and on and on, and, and this is the key, thisis where the limit of the Limit Comparison Test comes into play, and, if the limit, the limitas n approaches infinity, of a sub n over b sub n, b sub n is positive and finite,is positive and finite, that either both series converge, or both series diverge,so let me write that. So then, that tells us that either, either both series converge, or both diverge, whichis really really useful. It's kind of a moreformal way of saying that, 'Hey look, as n approachesinfinity, if these have similar 'behaviors then they are either going to 'converge, or they'reboth going to diverge.' Let's apply that right over here. Well if we say that our b sub n is one over two to the n, just like we did up there, one over two to the n, so we're going to compare, so these two series right overhere, notice it satisfies all of these constraints,so let's take the limit, the limit as n approaches infinity of a sub over b sub n, so it's going to be one over two to the n minus one over one over two to the n, andwhat's that going to be equal to? Well that's going to be equal to the limit as n approaches infinity, of two, if you divide byone over two to the n that's just going to be thesame thing as multiplying by two to the n, so it's going to be two to the n over two to the n minus one. Over two to the n minusone, and this clearly, what's happening in thenumerator and the denominator, these are approaching thesame, well we can even write it like, we caneven write it like this, divide the numerator anddenomiator by two to the n, if you want, although it's probably going to jump out at you at this point. So, limit as n approaches infinity, let me just scroll overto the right a little bit. If I divide the numerator by two to the n, I'm just going to have one. If I divide the denominatorby two to the n, I'm going to have one minusone over two to the n. And now it becomes clear,this thing right over here is just going to go to zero, and you're going to have one over one. The important thing is that this limit is positive andfinite because this thing is so this thing right overhere, is positive and finite, the limit is one is positive and finite, if this thing converges andthis thing converges, or this thing diverges, and this diverges, well, we already knowthis thing converges, it's a geometric series where the common ratio is less thanone, and so therefore this must converge as well,so that converges as well.
