 Two of the more important convergence tests are based on geometric series. So if we take a simple example, we know that in the geometric series r to power n, the nth term is r to the n. So the nth root of the nth term of the geometric series is r. This earth-shattering revelation is most useful because we know when a geometric series converges. If our sum is a geometric series, then we know that our series diverges if the absolute value of r is greater than or equal to 1. And we also know that the series converges if the absolute value of r is less than 1. And this suggests the following idea. If some series is like a geometric series, then the limit as n goes to infinity of the nth roots of the terms should be equal to r. And so this suggests the root test. Suppose we have a series where the limit as n goes to infinity of the nth root of the absolute values of the term is equal to l. The series converges if l is less than 1, diverges if l is greater than 1, and if l equals 1, the test is inconclusive. An easy way to remember the root test is that if our limit as n goes to infinity of the nth roots of our terms is equal to l. Then our series is like a geometric series with common ratio l. So let's try and determine the convergence or divergence of the series whose terms look like 5 to the n over 3 plus n to the n. So the first important question to ask yourself is which convergence test will we use? And in this case, we know to use the nth root test because taking the nth root of an nth power seems to be an easy thing to do. So we want to find the limit as n goes to infinity of the nth root of the absolute value of our terms. Now for any positive whole number value of n, 5 to the n over 3 plus n to the n is going to be a positive number. So we don't need the absolute value operators here, and we can ignore them. And since I'm taking the nth root of an nth power, those two operations are inverse operations. So I'll be left with the limit as n goes to infinity of 5 over 3 plus n, and I'll find that limit, which is 0, and since the nth root converges to a number less than 1, the series converges. We might take a look at a geometric series in a different way. We know that in the geometric series the ratio of consecutive terms is r. So the limit of the ratio of consecutive terms is r, and again we know the series diverges if the absolute value of r is greater than or equal to 1, and the series converges if the absolute value of r is less than 1. And so this suggests that if we have some series that's like a geometric series, the limit as n goes to infinity of the ratio of successive terms is going to be some ratio r, and the series will converge if the absolute value of r is strictly less than 1. And this suggests what's known as the ratio test. Suppose I have a series where the limit as n goes to infinity of the ratio of successive terms is equal to some value l. Then the series converges if l is less than 1 and diverges if l is greater than 1. And again, if l is exactly equal to 1, the test is inconclusive. And again, a useful way of thinking about the ratio test is that if the limit of the ratio of successive terms is l, then our series acts like a geometric series with common ratio l. So let's try and determine the convergence or divergence of the series 5 to the n over n factorial. We'll use the ratio test because the quotient of successive factorials will give a simple expression. So we want to find the ratio between one term, 5 to the n over n factorial, and the next term, 5 to the n plus 1 over n plus 1 factorial. Again, for any positive whole number value of n, all of these terms are going to be positive numbers so we can drop the absolute value operators and simplify our expression. Let's do a little bit more algebra. 5 to the n plus 1 is 5 to the n times 5. Meanwhile, remember that n factorial is the product of the whole numbers from 1 to n. So we note that n factorial is the product n times n minus 1 times n minus 2 all the way down to 3 times 2 times 1. And n plus 1 factorial is the product n plus 1 times n times n minus 1 and so on, or it's n plus 1 times n factorial. So we can replace n plus 1 factorial with n plus 1 times n factorial. We can simplify this expression and take the limit as n goes to infinity, and since the ratio of successive terms converges to a number less than 1, our series converges.