 In this video, we'll use the integral test to identify whether the harmonic series converges or diverges, and when a p-series converges and diverges. We'll begin with the harmonic series and the associated integral from 1 to infinity of 1 over x dx. In evaluating this integral, we find that it's equal to the limit as b approaches infinity of the natural log of absolute value of x evaluated from 1 to b, which is the limit as b approaches infinity of natural log of absolute value of b minus natural log of 1. Now as b approaches infinity, natural log of b will also approach infinity. So this integral diverges. Since this integral diverges, then the series itself must diverge as well according to the integral test. So the harmonic series diverges. Now consider the p-series for some constant p is greater than zero. The series is defined as k equals 1 to infinity of the sum of 1 over k to the p, where p is some constant greater than zero. And it's associated integral from 1 to infinity of 1 over x to the p dx. Now evaluating this integral, we find this is, I can rewrite this as the integral from 1 to infinity of x to the negative p dx, the antiderivative of which is 1 over negative p plus 1 x to the negative p plus 1. And since this is an improper integral, we can rewrite this as the limit as b approaches infinity of this evaluated from 1 to b. Using the fundamental theorem of calculus, we find this as the limit as b approaches infinity of 1 over negative p plus 1 times b to the negative p plus 1 minus 1 over negative p plus 1 times 1 to the negative p plus 1. If we clean this up a bit, we find this as the limit as b approaches infinity of 1 over negative p plus 1 b to the negative p plus 1 minus 1 over negative p plus 1. Now as messy as this looks, let's actually consider the two cases that would make the difference here. Case one, when p is greater than 1, if p is greater than 1, then we know that negative p plus 1 is less than zero, which tells us that the limit as b approaches infinity of b to the negative p plus 1 is equal to zero. So the integral must converge. According to the integral test, this tells us that the series converges for p greater than 1. Now suppose that p is between zero and one. If so, then we know that in this case, negative p plus 1 is greater than zero, which tells us that the limit as b approaches infinity of b to the negative p plus 1 is equal to infinity. So the integral in this case must diverge. According to the integral test, this then tells us that the series diverges as well. So in summary, we find that for any series of the form, k equals 1 to infinity of 1 over k to the p, where p is greater than zero, the series converges when p is greater than 1 and diverges when p is between zero and one.