So when doing this problem, we're going to figure out what two terms would add to give me that 1 / n + 1 and plus two term. So let's make that into a / n + 1 and b / n + 2. So we'd have a * n + 2 + b * n + 1 equaling one if we let. N equal -2 we'd have this first term canceling, so we'd get one equaling b * -2 + 1, so B would be -1. If we let N equal -1, we'd have one equaling a * -1 + 2 and the second term would cancel, so we could see that a equal 1. So a different way to write this would be a summation of N equal 1 to Infinity of 1 / n + 1 - 1 / n + 2. Now let's make sure that I'm starting my index in the right spot. If I have one here, I get 1 / 1 + 1 is 2 - 1 / 1 + 2 is 3. So if I have 1/2 - 1/3. We'd get 3 - 2 / 2 * 3, which is our first term there, so we're OK with that. Now we're going to look and realize by putting in a couple terms. If I get 1 / 1 + 1 - 1 / 1 + 2 + 1 / 2 + 1 - 1 / 2 + 2 plus... This is going to be 1/2 -, 1/3 + 1/3 - 1/4. So it's going to be a telescoping series. And we realized that the second term and the first term of each of these parentheses are going to cancel. So this is going to really equal 1/2 - 1 / n + 2. So now what we need is we need to figure out what happens when that N gets really, really large. So if N is going to Infinity, we have one half 1 / n + 2. So that n + 2 is going to get really, really big. But one over really, really big is really, really small. So it's going to go to 1/2.