 In this video, we're gonna discuss the solution to question 14 from the practice final exam for math 12-20, in which case we're asked to consider the series where we take the sum where k equals two to infinity of one over k times the natural log of k squared. And that's to say the natural log of k, that whole quantity of squared right there. There's two parts of this question. This question is gonna ask us to do some remainder estimates, some error estimates associated to approximating integrals, sorry, approximating series with respect to partial sums. And it's broken up in two parts. The first part you can see on the screen right now, which actually has a statement right here. The series can be found convergent by the integral test. So this is not something that you have to prove. This is given to us that the integral test will provide that this thing is convergent. Now you are asked to verify by computing the following formula, which be aware that again, you're not gonna cite the integral test. This is already going on here, but if one was using the integral test, you'd have to compute essentially this integral right here. Now there is one important difference I wanna mention that the series starts off at k equals two, but the integral is gonna start off with n. So we're having a placeholder right here, which we could easily insert two into that to determine the convergence by the integral test. Now in order to calculate this improper integral, I would recommend using the u substitution, take u to be the natural log of x, and therefore du equals one over x dx. And you'll notice that the dx of course is there and then the one over x is there as well. And so then this would translate as we switch the variable here, you're gonna get a du over u squared. I also wanna change the bounds, right? As you plug in into this formula, you're gonna get the natural log of n as your lower bound. And as we do infinity right here, treating this as a limit, we take the natural log of infinity that is take the limit as x approaches infinity of the natural log of x, and that itself will become infinity. So we still have this improper integral. Now integrating this thing right here, well by the power rule, since we have a one over u squared, we can treat that as u to the negative two that would raise the power by one to be u to the negative one that is one over u. And then divide by that exponent, which is a negative one, you get a negative one over u for the anti-derivative and evaluate this from negative natural log of n to infinity. Now because of the negative sign, I actually wanna switch it to be a positive, so I'm gonna switch the order of the limits there. And so we get one over u as you evaluate from infinity to the natural log of n. And then as you evaluate these expressions right here, you're gonna get one over the natural log of n minus one over infinity. Of course, one over infinity is to be interpreted as a limit. We're taking the limit as u approaches infinity of one over u, and that'll go off towards zero. And so this integral is equal to one over the natural log of n, and this is what part a is asking us to do. Now going back to this idea with the integral test, if you plugged in n equals two into this formula, you would see that this is a finite number. So therefore the integral is convergent, which implies the series is convergent via this integral test. So then look at part b here. What is the smallest choice for n that would guarantee that the partial sum where k raised us from two to n of one over k times the natural log of k squared? What would be the smallest choice of n to guarantee that this partial sum would be an accurate approximation of the series to within one over the natural log of 100, right? And so this is a statement about error bounds right here. So this is some prescribed error, and we need to find that our remainder, rn of x needs to be less than this. Now the important thing to remember about the integral test, and this is what we mentioned earlier, is that the remainder of the integral test will be bounded above by the integral from n to infinity of the function one over x times the natural log of x squared dx. And as we showed in part a, this is equal to one over the natural log of n, right? And so this is supposed to be less than or equal to one over the natural log of 100. And so this inequality right here we need to solve. Well, taking reciprocals, we see that the natural log of n must be greater than or equal to the natural log of 100. The inequality switches direction when you take reciprocals, and then exponentially able sides, which is an increasing function that doesn't change the direction of the inequality, we're gonna get that n needs to be greater than or equal to 100. So 100 is the smallest number that we can guarantee that this thing will be accurate to this prescribed amount of error. And so in this exercise what we did here is we, on part a, we computed the error bound associated to the series, and then on part b, we used it to determine how accurate, how many ends do we need in order to guarantee accurate to a certain level. And so you'll see something like this very similar on the final exam. Of course, it might not be an error bound associated to the integral test. It could be some of the other error bounds that we've seen this semester. So be prepared to answer any of those.