 This lesson is on the ratio test and the alternating series test, and then we will go to an alternating series estimation theorem. The ratio test is the last test that you will have on positive series. The alternating series, of course, will alternate signs. The terms will be both positive and negative. So let's look at the ratio test, the definition of the ratio test. Let a sub n be a series such that the limit as n approaches infinity of the absolute value of a sub n plus 1 over a sub n is equal to some limiting value. One of the following things can happen. The L can be greater than 1, and that means your series will diverge. If L is less than 1, then the series converges. If L equals 1, this particular test doesn't tell us anything at all. So we have those three conditions that can happen. Let's look at problems dealing with the ratio test. The first one says n squared over 2 to the n. So let's do the limit as n approaches infinity of a sub n plus 1, which means everywhere you see an n, you put an n plus 1 quantity squared over 2 to the n plus 1 times 2 to the n over n squared. And instead of putting one over the other, when you have these fractions, what you do is just take the one without the n plus 1, which is the plain old n, and flip it over. Very convenient way to look at this. Let's go on and simplify this limit as n approaches infinity. Well, we have n plus 1 squared over n squared, and the limit as n approaches infinity of that. If you remember back at Lobotel's rules and limits and all of those things, that actually is a 1. So we won't have to worry about it. So we just have to work with 2 to the n over 2 to the n plus 1, which is 1 over 2. So this limit is 1 half. And 1 half, of course, is less than 1, which means our series converges. Let's go on. What about 2 to the n over n factorial? Now the ratio test works very, very well with factorials. So let's try this one. Limit as n approaches infinity of 2 to the n plus 1 over n plus 1 factorial times n factorial over 2 to the nth power. Simplify this limit as n approaches infinity. 2 to the n plus 1 is 2, n factorial over n plus 1 factorial. Well, n plus 1 factorial can be written n plus 1 times n factorial. So the n factorials will go out, and we'll have 2 over n plus 1. And that goes to 0, which is less than 1. Again, our series converges. Let's go on. n factorial over n plus 3 factorial. What happens here? Limit as n approaches infinity of n plus 1 factorial over n plus 4 factorial times n plus 3 factorial over n factorial. So this equals the limit as n approaches infinity of n plus 1 factorial over n factorial is n plus 1. n plus 3 factorial over n plus 4 factorial is also n plus 1. So this one actually equals 1. So this particular test, the ratio test tells us nothing about the convergence or divergence. Let's go on. This one says n factorial over e to the n. Just looking at this, one of the things we know is that factorials grow faster than any other type of series that we will work with, even faster than an exponential. So this one should diverge. So let's look at this. Limit as n approaches infinity of n plus 1 factorial over e to the n plus 1 times e to the n over n factorial. So that equals the limit as n approaches infinity of n plus 1 factorial over n factorial is n plus 1. Then we have e to the n over e to the n plus 1, which is e. And as n approaches infinity, this approaches infinity, which means the series diverges. Let's try another one. n over ln of n to the nth power. Try the ratio test on this. There is another test, by the way, that you can use, which is called a root test. And you will find that in several books. Most of the time the ratio test will cover anything you can do with a root test. And it is not presented in your particular book. So we have the limit as n approaches infinity of n plus 1 over ln of n plus 1 to the n plus 1 times ln of n to the nth power all over n. Now as n approaches infinity using your old rules of limits, we know that n plus 1 over n will give us a 1. Now what about this ln of n to the nth power over ln of n plus 1 to the n plus 1? Well, let's look at that limit on our calculator. And I've typed the function fully in here all the way to the end, to the x plus 1. And of course x is your variable when you're doing limits. One of the nice things about this calculator, it does limits very nicely. And then we can even limit to infinity and get an answer on that. And the answer is zero. So our answer here is zero, which is less than one, which means our series converges. Let's try one more. How about n times 3 to the n over 4 to the n minus 1? Again, use that ratio test. It's a very formidable test. 3 to the n plus 1 over 4, n minus 1 plus 1 just gives us 4 to the nth power. And this one's 4 to the n minus 1 over n times 3 to the nth power. Cleaning this up. Again, we can look at the n plus 1 and n and make that into just plain old 1. And then we have 3 to the n plus 1 over 3 to the n, so that gives us 3 in the numerator. 4 to the n minus 1 over 4 to the n, which gives us a 1 over 4 in the denominator. And if we evaluate 3 fourths, we get 3 fourths. So actually this one's less than 1 and our series converges. And that seems very surprising. We do have our geometric series here, but it's multiplied by n. And we find just by doing this that it does certainly converge. Let's go on to alternating series. This is a very interesting test. First of all, we have to have a series that alternates with signs, which means it goes from positive to negative to positive back and forth. You do not have to have that negative 1 with a power sitting in there. It could be something like a cosine and a sine where one is positive and one is negative. Just watch for the alternating piece. And when we talk about a sub n in this one, we do not mean the negative 1 or the alternating piece. We just mean that positive piece. In an alternating series, the series converges if the following conditions hold. All of these have to hold. The a sub n's are all positive. a sub n is greater than a sub n plus 1, which means the terms in the sequence on that series go smaller and smaller. And last but not least, the limit as n approaches infinity of a sub n equals 0. And of course, that is very, very important for a series to converge at all. So let's try some problems on the alternating series. Determine the convergence of negative 1 to the n over n to the fourth. Well, we just have to follow three rules here. A, does it have positive terms? Well, 1 over n to the fourth is certainly positive all the way. B is a sub n plus 1 less than a sub n. Well, if we put in 1 over n plus 1 to the fourth, that certainly is smaller than 1 over n to the fourth. And part C is the limit as n approaches infinity 0. And we just have to test it out, and it certainly is 0. Now the other thing we look at with this, we not only say this converges, but we have two things we look at. Converges absolutely, converges conditionally. This converges according to the alternating series test. But does it converge according to some of our other tests or any of our other tests? And the answer is yes. And if the answer is yes, then we say this alternating series converges absolutely. So when we have absolute convergence, that means our series not only converges with the alternating series test, but with another test. Let's do another one. Determine the convergence of negative 1 to the n over n. So let's do the A, B, C again. So A, are the terms positive? Yes. B is a sub n plus 1 less than a sub n? Yes. It's getting smaller and smaller. And C is the limit going to infinity 0. And we can say yes. So the series converges. But let's look at the normal series without the alternating piece, which is 1 over n. Does this series converge? Well, this is your harmonic series where p is equal to 1. So this one we know diverges. So because your non-alternating series diverges and your alternating series converges, we say that this converges conditionally. Absolutely for the ones that do the double, conditionally for the ones that only converge through the alternating series. Let's go on and try another problem. How about negative 1 to the n over the square root of n? Does it follow A that the terms are positive? Yes. B that they're decreasing? Yes. Is the limit 0? Yes. How does the series 1 over the square root of n converge? Well, it doesn't. So because this one does not converge, again, we say our alternating series converges conditionally. Let's try another one. Looks like another p series, but this time we stuck in arc tangent. So what happens when we put that arc tangent in? Well, does it follow A? Well, as n approaches infinity and we only look at arc tangent, the arc tangent of infinity is approximately pi over 2. So it is a number. So it is a positive number. So are the terms positive? Yes. Do they decrease? Well, the n to the fourth power is telling us yes they do. And last but not least, does the limit approach 0? Yes. Now, what about the series arc tan of n over n to the fourth? Arc tangent of n, as n approaches infinity, is a number. So it's a small number over this very large n to the fourth number. So this series did converge originally. So we can say that our alternating series converges absolutely. Let's try another one. This one reads negative 1 to the n plus 1 times n squared times 2 to the n over n factorial. A, are all the terms positive? Indeed, yes. B, are they getting smaller and smaller? Even though we have 2 to the n here which grows exponentially, n factorial grows larger. This is the largest discrete math growth that we have. So the n plus 1 is larger. So it is decreasing. And C, it does go to 0. So it converges through the alternating series test. But let's see if it converges with any other test. And the test we use is the ratio test. So we take the limit as n approaches infinity of n plus 1 squared times 2 to the n plus 1 over n plus 1 factorial times n factorial over n squared times 2 to the n. Let's clean this up a little bit. We have n plus 1 squared and n squared can't do too much with that right now. 2 to the n plus 1, 2 to the n leaves us in 2 up here. n factorial over n plus 1 factorial leaves us in n plus 1 in that denominator. So how can we reduce this? Well, as n approaches infinity, if you remember, n squared over n squared is going to give us a 1. So we end up with 2 over n plus 1 and that is going to approach 0. And since 0 is less than 1, this converges. So we can say that our original series converges absolutely. Let's try one more. Negative 1 to the n times n plus 1 over n squared. A are all the terms positive? Yes. Does this decrease? Yes. And part C does it approach 0 and we know that's ES2. What about converging absolutely versus conditionally? Well, we are looking at the series n plus 1 over n squared and we know this approximates series 1 over n. We can use a comparison test to determine that. And of course, this series does not converge. Again, it's our harmonic series. So the original series converges conditionally. Now what about this alternating series estimation theorem? This is a really interesting theorem. It says if you have a convergent alternating series and you take a partial sum, then the remainder on that partial sum satisfies that s minus s sub n is greater than or equal to the remainder that's less than or equal to a sub n plus 1. And what this tells us is that the remainder is telling us how far away we are from the actual answer on the summation. Let's try a problem. Let's say we're given negative 1 to the n plus 1 times 1 over n and we know this actually has some sort of a sum to it. So we're asked is s sub 100 or s sub 100 an overestimate or an underestimate of s. Well, let's try summing the series. And again, we can go to our calculator and do this particular problem. Let's go to catalog and go to sigma or sum. And this says expression variable low high. So we want to put our expression in first. So we have negative 1 to the n. So I'm going to use x in this times 1 over x. That's our expression. There's our variable and we're going to go 1 comma infinity with the green. And let's see if it can compute anything for us. It doesn't quite give us what we want. So let's just try another number. Let's go back to and go change our number to 500. So it's approximately negative 0.6921. Okay, when we sum this up to 100, we get negative 0.688172. Is this an overestimate or an underestimate? Well, we can see that it is an overestimate. Then the next question is by how much? Well, the nice part about this is we've summed to 100. We only have to find out what a sub n plus 1 is because we have a sub n being the hundredth term or a sub 100. So we just have to find out what a sub 101 is and we can find out by how much this is an overestimate or an underestimate. And if we put 101 in our formula, we have 1 over 101, which is approximately equal to 0.009901. So we are very, very close to the final answer according to what we see here. And I only went out to 500, so this is still very, very close to our final answer. Our last question is how many terms must be summed for the estimate to be within 1 1000 of the actual sum? Well, remember our sigma is negative 1 to the n plus 1 times 1 over n. This is very convenient. So to be within 1 1000, our a sub n plus 1 term has to be equal to 1 over 1,000. So the term before it is the number we need summed up. So our n has to equal 999. This concludes your lesson on ratio tests and alternating series tests.