 This lesson is on series or tests for convergence. In the last lessons you worked with series, but you actually found the sums of convergence. This starts your lessons on only the tests. Now we have several tests, but we will have two that we will introduce today. The first one is the nth term test for divergence, and the second one is the integral test. And with those two tests, we'll talk about the p-series. So let's go on. The first one is the nth term test for divergence. The reading on this says if the limit as n approaches infinity of a sub n does not equal 0, or if the limit as n approaches infinity of a sub n does not exist, then the sum from n equals 1 to infinity of a sub n is divergent. So let's look at example one. Using the nth term test for divergence to determine whether a series is divergent. The test tells you nothing about convergency. Just remember that. It only talks about divergency. So let's look at our first example. The sum when n equals 1 to infinity of n squared. Well, if we take the limit as n approaches infinity of n squared, that surely goes out to infinity. And we know this is not a decreasing function also. So this definitely is divergent. Let's go on. How about the sum when n equals 1 to infinity of n plus 1 over n? Well, we can break this one up as we put that limit as to 1 plus 1 over n. Well, as n approaches infinity, the 1 over n becomes close to 0. We're left with an answer of 1. The nth term test for divergence says that if the limit approaches some number or is another number other than 0, then the series is divergent. That means this is divergent. Let's go on. Using the nth term test for divergence, try this one. Sum from n equals 1 to infinity of 1 over n. If we take the limit as n approaches infinity of 1 over n, we get 0. Remember, the nth term test said it does not equal 0. So if it equals 0, we really cannot tell anything. So we cannot tell from this test. We will have other tests to tell us what is happening. Let's go on. Suppose we have negative 1 to the n plus 1 that we are summing. So if we take the limit as n approaches infinity of negative 1 to the n plus 1, well, if we start out with a 1, this will be squared. So we're going to have a 1. If we put in a 2, it will be cubed. So it will be minus 1 plus 1 minus 1 plus 1. They actually do not cancel out. So we do not know what the answer is because we could have a negative 1 at the end or we could have a positive 1 at the end and it won't cancel out. So this one too is divergent. Let's go on. Suppose we have the sum from n equals 1 to infinity of 1 over the square root of n. We'll take the limit as n approaches infinity of this one and we see that that goes to 0. Again, we cannot tell. Another example, what about 1 over n cubed? Again, as n approaches infinity, we can't tell because the limit as n approaches infinity of 1 over n cubed is equal to 0. Again, we can't tell. What about 3 over 2 to the n? Again, when we take that limit as n approaches infinity of 3 over 2 to the n, it equals 0 cannot tell. A lot of times you cannot tell. It doesn't say that this is convergent. Now you already know that this particular one is convergent because it is a geometric series. What about 3 times negative 1 to the n over n squared? Again, when we take the limit on this as n approaches infinity, we get 3 times negative 1 to the n and of course that will be an alternating series and then over n squared and as n approaches infinity then this whole fraction becomes 0. So again, we cannot tell. So more times than not, the n-term test for divergence doesn't really help us but it does help us in those few cases that we can just take that limit and determine that the series is divergent. This is the integral test and it reads, let a sub n which is the terms in our series equals some sort of a function f of n and our f of x that we turn this into is a continuous positive decreasing function where x is greater than or equal to some constant. Then if the integral from c to infinity of f of x dx converges, then our series from n equals 1 to infinity of a sub n converges and if the integral from c to infinity of f of x dx diverges, then our series, the sum from n equals 1 to infinity of a sub n does diverge. So what we are saying here, if we can take the integral of the terms in our series and if it converges to some point through improper integrals of course, then the series converges and if that integral diverges and does not get some number, then the series diverges. Now you might think this works for every single series that you've ever worked on but remember we can't integrate every function we have ever seen. Let's go on and see how this works. So let's look at the sum when n equals 1 to infinity of 1 over n. Let's test this one. Put that in integral form and we get integral from 1 to infinity of 1 over x dx. When we integrate that, we get ln of the absolute value of x from 1 to b, remember improper integrals and we take the limit as b approaches infinity. And we see when we put in ln of the absolute value of b minus the ln of the absolute value of 1, ln of the absolute value of 1 is 0 and as b approaches infinity, ln of b is infinite, so this one is divergent. Therefore, the series of 1 over n is divergent. And we'll talk about this one in a few minutes with p-series. Let's go on. 1 over n squared. Well what happens when we integrate this one? We get the integral from 1 to infinity of 1 over x squared dx. We all know that that equals negative 1 over x from 1 to b limit as b approaches infinity. And that becomes negative 1 over b minus 1 over 1 and a limit as b approaches infinity and this answer becomes 1. We do get a result. Therefore, our series is convergent. Let's try another one. Most of these that I am talking about is under the title of p-series. Again, integrate this one from 1 to infinity. 1 over the square root of x dx and that becomes x to the 1 half times 2 and we're going to do the limit as b approaches infinity of 2x to the 1 half from 1 to b. By substituting b at infinity, this will become infinite. So again, this one is divergent. Moving on. What about something like 2n plus 1 over n squared plus n minus 2? Does this one integrate? Well, let's see. Integral from 1 to infinity of 2x plus 1 over x squared plus x minus 2 dx. And we'll see if we make u equal to x squared plus x minus 2 and du equal to 2x plus 1. We certainly have something that we can integrate. So let's go on. This becomes du over u, which leads to ln of the absolute value of u. And if we go back to u, which is x squared plus x minus 2, we can put in ln of the absolute value of x squared plus x minus 2. And we are taking this from 1 to b where limit as b approaches infinity. If we put infinitely large numbers into our x squared plus x minus 2, we see that this comes out to infinity. So again, we can say this is divergent, which means our series also is divergent. Anytime we can take the integral, we can use the integral test. So just watch out because this is a very good test. Now, we are not getting the sum of the series. We are just determining whether the series is convergent or divergent. Let's talk about p-series and harmonic series. What is a p-series? A p-series is when we have the sum of n equals 1 to infinity of 1 over n to the p. And we have done a few of these the last few minutes. And p is a positive number. What happens? Well, if p is greater than 0 and less than or equal to 1, then the series is divergent. And we saw that with 1 over n, 1 over the square root of n. Now, if p is greater than 1, then the series is convergent. So in our p-series, the 1 is the important 1. And if we have the sum of n equals 1 to infinity of 1 over n, that is called the harmonic series, very important series. So one more time, if p is between 0 and 1, our series diverges. If p is greater than 1, like 1 over n squared, then the series converges. And our harmonic series is called the 1 over n series. This concludes your lesson on series, test for convergence, p-series, integral test, and the term test for divergence.