 Welcome back everyone. Now we're ready to talk about the ratio test of all of the convergence tests that we've computed in this lecture series I would say that the ratio test is probably my favorite of all of them So it's sort of like the best for less right here and the ratio test gets its name for the following reason I want you to consider the sequence of ratios. So you have some sequence a sub n like you see right here I want you to take the ratio of consecutive terms So take the term a n and divide it from its successor a n plus 1 or if you want to think of it take Take a term in the sequence and divided by its predecessor So take the ratio sequence of consecutive terms also take absolute value because we don't really care about positive or negative right here and So take the limit of the ratio So take the limit as n goes to infinity of a n plus 1 over a n all inside absolute value Let's say that this sequence is convergence. So the sequence of ratios is converges to some limit value call it L So that's how the ratio test gets its name. We have to compute the limit of this ratio Now if that limit is less than 1 then we can conclude by the ratio test that the series n equals 1 to infinity the sum of a sub n is Absolutely convergent now remember what absolute convergence here means it means that if you take the sum of Absolute values, that's a convergent series But as we've seen with the test of absolute convergence that if a series is absolute convergent then it's can its convergence as well And so the ratio test doesn't just give us convergence It actually gives us absolute convergence if this limit is less than 1 Now be aware that we get we can get away with absolute absolute convergence here because again We're kind of taking absolute values It's built into the mechanism of the ratio test here and this limit right here is basically measuring how close together Are the terms in the sequence how rapidly is the sequence of ratios shrinking because if this thing is shrinking Rapidly enough then its limit will be something less than one I should say that if the terms in the sequence are close enough together and That's going to give us that this sum right here will be convergent because the terms a n are shrinking at a wrap-enough rate That's that's what we're going to get from this limit right here Now on the other hand if the limit is greater than one if their limit of the ratios is greater than one Then this tells us that the series is in fact Divergent right it'll be a divergent series and the other possibility is if the limit of ratios is equal to 1 Then the ratio test is actually inconclusive it turns out in that situation the series could be absolutely convergent It could be convergent, but not absolutely that is it could be conditionally convergent or it actually could be divergent I'll show you some examples of that in just a second But let me show you some affirmative situations first so if we're going to test If we're going to we're going to try the testing the convergence of a series We'll use the ratio test here take the sum or n equals one to infinity of the alternating series negative one to the end You have n cubed over three to the end now you'll notice here some things going on here There's this geometric part three to the end, but this is not a geometric series There's this like in cube floating around, but this is not a p-series There's also this negative one to the end. It is actually an alternate series So one could try to determine whether this thing is convergent by the alternate series test Which you could be successful in doing that but one advantage of using the ratio test here is that the alternate series test can only show Convergence, but the ratio test can show Absolute convergence and then frankly it might actually be able to do it in an easier fashion So how does one actually work this thing out? We want to consider the the ratio sequence take a sub n divided by a n and so oftentimes what happens is that the terms in your sequence are actually fractions So when I actually work at this sometimes I actually want to take the a n plus one term and then kind of separate Separate it as a one over a n term because these terms are often both fractions themselves So the a n plus one is going to look like the following We have negative one to the n plus one power. We're gonna get in plus one Let me rewrite that we're gonna get in plus one cubed all over Three to the n plus one like so that's the first term and then the next term is gonna look like taking the reciprocal We get three in on top and then we're gonna get a negative one to the n times n cubed like so And so we want to simplify this thing That's the next thing you have to do is you have to simplify this expression now some things to mention here Is that since you're taking absolute values any factors of negative one can just be entirely ignored It does not matter whether or pods are negative those negative signs will always disappear with the absolute value The next thing I want to mention is that you want to rewrite this fraction putting things of similar type together Like notice how there's this exponential term three to the n and there's this three to the n plus one I want you to rewrite it so that those two friends are now together So we have three to the n over three to the n plus one and then the other terms We have this n plus one cubed and this n cubed those are kind of the same type put them together as your next fraction You're gonna get in plus one cubed over n cubed Like so and again now we're gonna try to simplify these things Because when you look at three in plus one on the bottom here This thing can break apart as three to the n times three and so this three the n cancels with that three to the End and you're left with the absolute value as one third, which is actually just a one third right here Now with the second bit in plus one cubed over n cubed both the numerator and denominator are being cubed And so if we bring out the exponential we're going to end up with n plus one over n cubed Are we writing it one more time we get one third we're going to get one plus one over n quantity cubed here And so in this situation as we take the limit as n goes to infinity because again That's what we're trying to do here. We want to take the limit All right, so this is a limit calculation the limit as n goes to infinity If you carry that through all of these calculations we get to the very end here We take the limit as n goes to infinity Well as n goes to infinity The one third will stay one third the plus one will stay there this one over n part will vanish to zero And so we end up with this this this sequence of ratios will converge towards one third times one cubed And which of course that would then just become the number one third like so This is the limit l that we saw at the ratio test. This is the limit of the ratios All right, and so because it's the limit of the ratios We want to compare with the ratio test the ratio test that we bring it back over here If our limit is less than one this says the series is absolutely convergent And that's exactly what we see over here This limit is less than one and so then we conclude by the ratio test That our series is absolutely Convergent and since it's absolutely convergent that actually implies it's convergent So if you're asked is the series convergent you would say yes It's convergent if you want if you could also further specify It's absolutely convergent and we use the ratio test in so doing As another example, let's take the let's take the sum of n equals 1 to infinity of n to the n over n factorial Now, this is not an alternating series So we couldn't use the alternating series test This in factorial makes it very difficult to integrate because what's the anti derivative of x factorial e? You know, that's that's kind of a weird thing that that's not a continuous function that we can find an anti derivative of But the ratio test actually works out really great in this situation Because our ratio test again, we're looking at the ratio of a n plus 1 over a n This is going to look like n plus 1 to the n plus 1 power over n plus 1 factorial That's the a n plus 1 term and then if you look at the 1 over a n term That's going to be in factorial over into the n And like we did last time put together similar types like we're going to put the factorials together And then put the exponential expressions together as well if we do the factorials We end up with n factorial over n plus 1 factorial And that's going to be paired up with the n plus 1 to the n plus 1 over n to the n like so All right Now much like we've done the previous example these ratios have to be simplified in some manner Now n plus 1 factorial has a very nice factorization as a factorial. This will factors n plus 1 times n factorial for which the n factorials then cancel right here And then another sort of curious thing here is that you have an n plus 1 The n plus 1 that's in the denominator can actually cancel with one of the n plus ones that are over here And so this ratio simplifies to be n plus 1 to the n above n to the n And much like our last example since we now have They're both to the power of n we can bring that out of our fraction and get n plus 1 over n To the n right here Uh as simplifying this we're going to get 1 plus 1 over n to the n now this one's a little bit more tricky um because we can't if we just set in to infinity right you're going to end up with uh this thing will look like This thing would look like 1 to the infinity which is actually in determined form So this kind of leads to a lopital argument For which if you want you can see the the video link that's in front of the screen right now You can actually see why this calculation turns out the way it is But this limit will turn out to be the number e Which is approximately 2.7, etc, etc This is our limit value and it is actually greater than 1 So what this tells us by the Ratio test the ratio test would give that this series is going to be divergent This is a divergent series by the ratio test And so when we go back to the ratio test we saw here if our limit's greater than 1 the series is divergent We're using that as we're using that evidence right here. Now I should mention that in this last example um, we found out that the limit The the limit of the an plus ones over an's to get absolute value there if this limit Goes off towards something some l value, which is bigger than 1 It turns out you actually didn't need to use the ratio test whatsoever you actually could have gotten away with the test for divergence or the so-called divergence test The test for divergence The reason that this limit goes to something bigger than 1 really comes from the fact that the sequence a sub n Does not go towards zero and if you look at the proof of the ratio test, that's why that's happening So we actually could have used the divergence test in that situation Because again going back to our sequence and play here this n to the n over in In the factorial here This isn't an obvious fact to us perhaps But the thing is the top is growing a faster rate than in factorial that this thing actually converges towards infinity Right. It's not the fastest growing function necessarily, but ultimately into the end The power exponential function beats the factorial so we could have used the divergence test there as well But you know, that's the thing about the ratio test is that if it turns out that you get in this situation You still get an answer, right? You don't just have to all your effort isn't wasted It's like I could have used the divergence test But since I didn't use it and I went to the ratio test I still get an answer as opposed to like the alternating series test Which it can tell you when the series is convergent But if if the alternating series test fails that doesn't imply convergence You'll need you actually have to do the divergence test in that situation I want to talk about at the last part of our video here. I want to talk about this last case a little bit. What happens? What about when the ratio test is inconclusive, right? Well, I mean if it's inconclusive that means you need to do something else But why do you need to do something else here? If it's inconclusive? We can draw no conclusions because of the following type of situations If you take if you take a first example here Take the sum of 1 over n squared, right in this situation This series is convergent. It's a p-series. In fact, it's absolutely convergent Because it's just a positive series there. That's convergent But if you look at the the ratio sequence a n plus 1 over a n This thing is going to look like n plus 1 squared over n squared and that all converge towards the number one. It's an absolutely convergent series, but um But the limit goes to one so if that limit ratio limit equals one it could be that it's uh, Absolutely convergent. It could be another option Take something like the following take the harmonic series 1 over n In this situation, we know that the harmonic series is divergent But if you look at the ratio sequence a n plus 1 over a n This thing that's going to look like a n plus 1 over n Which is that likewise converges to one so the limit when it's one it could be absolutely convergent, which is convergent But it could also be divergent and it turns out there is a third possibility, right that If you take the sum of like say the alternating harmonic series a n A negative 1 to the n plus 1 times 1 over n. This is actually an example of a conditionally convergent series But in that situation if you take a n plus 1 over a n You still end up with Just the same thing you did here You're going to end up with a n plus 1 over n because the absolute value is just negate the alternating fact This is going to go to one So you can see that when When this limit of ratios equals one we truly are inconclusive because you can absolutely convergent You can get divergent and you can get conditionally convergent Now one thing I should mention is that the conditionally convergent situation must occur in this case where the limit goes off to one But we could actually determine divergence or absolute convergence if the limit is greater than or less than one depends on the situation But be aware that if the limit of the ratios goes to one you have to use a different Test than the ratio test. So although I think the ratio test is super awesome It does of course have some limitations Particularly this this inconclusive case when the limit goes to one