 In this video we provide the solution to question number 11 for practice exam number 4 for Math 1220. We're given the series where we take the sum where k equals 2 to infinity of the sequence 1 over k times the natural log of k squared. And there's two parts to this problem here. So the first part is that the series can be, so just reading the problem here, the series can be found convergent using the integral test. And so we're going to verify this by computing the integral from n to infinity of 1 over x times the natural log squared of x dx. So notice that to show that this thing was convergent, we'd show that this integral is convergent specifically when you have a 2 right here. But we're going to do this in more generality, we're going to do it with an n. So that's where we want to compute this integral. Okay, so the integral from n to infinity here, 1 over x times the natural log squared of x dx. A u substitution is very relevant here. You could take u to be the natural log of x. So then du would equal dx over x for which then our integral would then become, you're going to get the 1 over x dx is going to become your du. And then the denominator, you have 2 natural logs, so you get a u squared. And so which case this is in the natural log of u. Excuse me, when you plug in n inside here, you're going to get natural log of n. So u is the natural log of n on the bottom and then at the top it's still infinity. As x approaches infinity, u will approach infinity as well. So using the power rule here, the anti-derivative is going to be negative 1 over u as you integrate from natural log of n to infinity here. Because of the negative sign, I'm going to switch the bounds around. This is going to be 1 over u as you go from the natural log of n on top to infinity on the bottom. And so then plugging that in here, you're going to get 1 over the natural log of n minus 1 over the natural log of infinity. For which the natural log of infinity is itself infinite. So you get 1 over the natural log of n minus 1 over infinity. That last term is then 0. And then simplifying, you end up with 1 over the natural log of n. And as n is any positive integer, this is going to be a positive number. And since this is not infinity, that would then show that this integral is a improper or this improper integral always converges. And so that's how you'd use the integral test. So because this integral is convergent, this series is convergent as well. Now we integrated this from n to infinity as opposed to 2 because we're not just after the convergence. We're also after the error bound. So this right here gives us an error bound with regard to the integral test here as we can use this to approximate things. So this gets us to the second part of the problem. What is the smallest choice of n that would guarantee that the partial sum where you take the sum where k equals n to k equals 2 to n of that same sequence 1 over k times the natural log of k squared there would be how what n guarantees that the partial sum approximation is accurate to within 1 over the natural log 100. For which you can use a calculator to approximate that, but I chose a number here specifically so you can do this without a calculator. So even if you forgot your scientific calculator, we can do this thing. So using the error bound with regard to the integral test, the error of truncating after the nth term is bounded by honestly the term we just found. It's going to be bounded above by this, which is that integral. So the error we want is the following. We take the error bound we found, which is 1 over the natural log of n. This needs to be less than or equal to 1 over the natural log of 100, like so. Taking the reciprocals, we need that the natural log of n. We need that to be greater than or equal to the natural log of 100. Taking the exponential of both sides, we need that n needs to be greater than or equal to 100. And so if we take the first 100 terms of the sum and add those together, that will then be accurate within this error that's described here.