 This lesson is on more tests for convergence. You do have several that you need to know. This lesson actually addresses the comparison test and the limit comparison test. So let's go on and look at the limit comparison test first. The limit comparison test reads, given the series a sub n, where a sub n is greater than 0, and you think it's convergent, then you pick a series called c sub n and compare the two. And the way you compare the two is you take the limit as n approaches infinity of a sub n over c sub n. And if that is less than infinity, or if it reaches a number, then your series converges. However, if you want to do it the other way and you want to divergent series, and you know your series is divergent, then you look for a divergent series called d sub n. Again, a sub n and d sub n are all greater than 0. Then you take the limit of a sub n over d sub n, and if that is less than infinity or equals some number, then the series a sub n diverges. So let's use the limit comparison test. Example, the series n equals 1 to infinity of 1 over 10 to the n plus 2. Now, we know this series converges, primarily because it is a geometric series with just 2 added to that denominator. So what should we compare it to? The obvious sum would be 10 to the n. Compare it to a sum of 1 over 10 to the nth power. So we'll do the limit as n approaches infinity of 1 over 10 to the n plus 2 over 1 over 10 to the n. And if we compare these two and we go out to infinity, we know the plus 2 does not matter. So this is approximately equal to 1. That is not infinity. It is a nice number. Therefore, comparing it to a convergent series means that the original 1 is convergent. So the sigma is n equals 1 to infinity of 1 over 10 to the n plus 2 is convergent. Let's try another one. This time, we want 1 plus cosine n over n squared. What will we compare it to? Well, we know cosine oscillates between negative 1 and 1. It won't make much difference as we go out to infinity. So we want to compare this to 1 over n squared. So compare to sigma of 1 over n squared. So we take the limit as n approaches infinity of 1 plus cosine n over n squared all over 1 over n squared. If we find the limit of this, we find that again, it's approximately 1 plus some cosine of n, which goes from negative 1 to 1. So we could have plus the cosine n. And that actually goes from 0 to 2. Again, any number from 0 to 2, nice number. So we say that our series 1 plus cosine n over n squared converges. Let's try another one. ln of n over n, what will we compare this to? We know that 1 over n is divergent. And ln of n will certainly be divergent. So let's compare it to ln of n. So now we have limit as n approaches infinity of ln of n over n all over ln of n. And that is equal to the limit as n approaches infinity of 1 over n, and that approaches 0. So ln of n, we know, to be divergent. Therefore, our series is divergent, too. Let's try another one. How about 100 plus n over 1 plus n cubed? What would we compare it to? Well, the 100 and the 1 don't matter so much. So we really have n over n cubed, which is 1 over n squared. So let's compare it to 1 over n squared, which we know to be convergent. So we take the limit as n approaches infinity of 100 plus n over 1 plus n cubed over 1 over n squared. And again, as we rationalized earlier, the 100 and the 1 really don't matter as you go out to infinity. The n over n cubed becomes 1 over n squared, and you have 1 over n squared. So that's approximately equal to 1. So our series converges. Let's go on. This time we have 3n squared plus n over the square root of 2 plus n to the fifth. Again, we think through this one in a very logical manner. The 3n squared is the highest power. And again, we're thinking about going towards infinity. So we have the n squared over the square root of n to the fifth, which really is n to the 2.5 power. So if we kind of approximate this to look like 3n squared over n to the 2.5 power, and the one we do choose actually is 1 over the square root of n. Because if you subtract 2 from 2.5, you get 0.5. So that would be the closest one here. So let's do the limit as n approaches infinity of 3n squared plus n over the square root of 2 plus n to the fifth. And we're going to compare that to 1 over the square root of n. And we eventually get something that approximates the 3n squared over n to the 2.5, which is over 1 over n to the 0.5. And putting these together, it's approximately equal to 3, a nice number, not infinity. So since our original series diverges, our series that we're working on diverges 2. What about the arctangent of n over n cubed? Well, we know arctangent goes from negative pi over 2 to pi over 2 for its range. So that number is not a big player when we're comparing it to n cubed as it goes out towards infinity. So let's compare this to 1 over n cubed. So compare to 1 over n cubed. So we take the limit as n approaches infinity of arctangent of n over n cubed over 1 over n cubed. And that, of course, is equal to the limit as n approaches infinity of arctangent of n. As we go out towards positive infinity, this becomes pi over 2. So since 1 over n cubed is convergent, then our series that we're looking at is also convergent. These are a few examples of how to use the limit comparison test. It is a fairly simple test to use in comparison to doing what we are going to do next, which is just the comparison test. So let's go on. The comparison test. Given a series, a sub n, where a sub n is greater than 0, there is a convergent series, c sub n, where c sub n is greater than 0. And this convergent series has to be larger than a sub n. So if we can find a series which is larger than the one we are working on and is convergent, then we can say our original series is convergent. Likewise, if we want a sub n to be a divergent series, then we pick a d sub n series which is smaller than a sub n and compare a sub n to that. So this one is a little trickier in trying to figure out what to compare our series to. So let's look at some of those examples again. Well, the first one is 1 over 10 to the n plus 2. In the limit comparison test, we just compare it to 1 over 10 to the n. But this time, we have to look a little differently. And we have to find a series that is larger than 1 over 10 to the n plus 2 to compare it to. And this time, we are going to compare it to 1 over 9 to the nth power, which we know is convergent. So just let's look at this on our calculators. I'm using my TI-83 for this because it is easier to see these functions as they go out towards infinity on the function mode rather than on the sequence mode. So I typed in 1 over 10 to the x plus 2. And I'm going to first compare it to 1 over 10 to the x, which is logical, but let's see what happens. I'm going to the table. And if you notice, my x is there at 43 and 44, which is fairly large. And I'm making this comparison. And you can see y sub 1 and y sub 2 are the same. We really want y sub 2 to be greater than y sub 1. So let's put in 1 over 9 to the x power. And look at that table this time. And we see, indeed, y sub 2 is greater than y sub 1. So 1 over 9 to the nth power would be the 1 to compare it to. We've compared to a series that is convergent. So our series, therefore, because we knew that from the beginning, is also convergent. Let's look at another one. 1 plus cosine n over n squared. What should we compare it to? Well, let's try comparing it to 1 over n squared again. So we want to compare to 1 over n squared. So let's look at the limit as n approaches infinity and see if 1 over n squared is larger than the 1 plus cosine and over n squared. And compare those on our calculator. In the beginning, you'll see something a little bit different. Let's go to our calculator on this and check the function against the comparison function, which is 1 over n squared. So we'll have 1 plus cosine x divided by x squared. And we'll compare it to 1 divided by x squared. And make sure we have all our parentheses correct and go to our table. And we see, indeed, y sub 2 is greater than y sub 1. So even as we go down to 49, and I'm sure as we go down further, we will see that, indeed, the y sub 2 is greater than the y sub 1. So 1 over n squared is the one to use. Since 1 over n squared, we know, is convergent. That's the one we chose. That means that the original one is also convergent. There are lots of choices here. We could have even picked 1 over n to the 3 halves power. Anything 1 over n to the 1.001 power. So as long as you compare it to a convergent series, you are safe on these. Let's look at another one. ln of n over n. Well, what should we compare this one to? Well, there are a couple of things. You can do the ln again, or we could just do 1 over n to some power. So let's try comparing it to 1 over n to the 3 fourth power, which we know to be divergent, because it's less than 1. So let's put those in our calculator. So we have ln of n, or ln of x in our case, divided by x, and then 1 to the 0.75 power. And again, let's look at that table. Table is easier on these than picking up a graph. And we see the one we're comparing it to is smaller than the original one, and that's what we want because we're looking for divergency. So indeed, this comparison works. So we can say because the one we're comparing it to is divergent, the original one is also divergent. OK, let's try another one. 100 plus n over 1 plus n cubed. This, again, is like 1 over n squared. So let's see when we do compare it to 1 over n squared. I'm using my TI-83 on here because it's a quick run to that table rather than a TI-89. It's easy to see this on the table, but either calculator will work on this, of course. If we compare it to 1 over x squared, that will be fine. Let's go to that table again. And we see that indeed, y sub 2 is less than y sub 1. And we are looking for this to be convergent. So we really want y sub 2 to be greater than y sub 1. So let's pick another value. So we have squared here. Let's pick something else. That one actually doesn't work. Let's try 1.5 this time and check that table out. And we see y sub 2 is larger than y sub 1, which means the one we're comparing it to is larger. And it is convergent. Therefore, our smaller one is also convergent. Obviously, we knew this all along, but this is actually using that test. So we compare to 1 over n to the 1.5, and this is convergent. Therefore, our series that we're working with is also convergent. Let's try the last one, octangent of n over n cubed. Let's compare it to 1 over n cubed. Now remember, we want 1 over n cubed to be larger than the octangent of n over n cubed. And this one has a possibility because anytime we put something up here, that will make this fraction smaller. So let's just check this one out to make sure that it actually works for us. So in our calculator, we'll put the original one and then the one we're comparing it to and go to our table and check this out. Well, it looks like the y sub 2 is smaller than the y sub 1, which means we really want it to be larger, so that one won't work very well. So let's try 2.5 and check that one out. So we see that our y sub 2, the one we're comparing it to is larger. Therefore, we can actually use that one for our comparison test. We could also use 1 over n squared, good candidate too. So let's change this to 2.5, which we know to be convergent. So that makes our series, octangent of n over n cubed to be convergent. Hope these have been enough examples for you to get started on comparison test and limit comparison test. Again, limit comparison test is a little bit easier to use and try to use that one as much as possible. This concludes your lesson on more tests for convergence and divergence.