If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:6:57

Worked example: Series estimation with integrals

Video transcript

- [Voiceover] In the last video we saw that if we have an infinite series where each term is a function of N, and the function itself is a continuous, positive, decreasing function over the intervals we care about, and we assume that this series converges, then we can estimate that series, we can estimate the value that it converges to, using a finite number of terms and some integrals, and the way that we establish these integrals is because when we split up the sum into finite sum and then another infinite series, we're about to conceptualize that infinite series in two different ways. It could be an underestimate of this integral right over here, or it could be an overestimate of this integral right over here, and because of that it allows us to not only estimate S but to put bounds on it. If we can evaluate the left hand side here and the right hand side here for a given infinite series, then we can establish our bounds on S, so let's see how good these are by applying it to a particular infinite series. So let's say that the infinite series we want to apply it to, let me do it in the yellow color, so it's going to be, let's say, we go from N equals one to infinity of one over N squared, and let's say we didn't know how to find the exact value here, but we want to estimate it. Let's say we want to estimate it using the first five terms, so we'll do S sub five is equal to the sum of N equals one to infinity, sorry, to five of one over n squared, and let's see, this is going to be equal to, let's get a calculator out, this is going to be one over one squared, which is just one, plus one over two squared, which is one fourth, plus one over three squared, which is one ninth, plus one over four squared, which is one 16th, plus one over five squared, which is going to be one 25th, and we get this value right over here, so I'll just say this is, I'll say it's approximately, one point four six four. Approximately one point four six four, and then if we evaluate each of these integrals, so our K in this case is going to be five. I just picked that arbitrarily. We could probably get a better, or we would get a better estimate if we had a higher K, if we did the first 10 terms or 20 terms, and we would have a worse estimate if we did K equals three, but I just picked K equals five because it seems reasonable, something that I could compute. I actually didn't need a calculator to do that. I could have done it by hand. It would have just taken more time. Let's evaluate these integrals here. So the integral from five plus one, so it's six to infinity of one over N squared DX, and, let's see, that's going to be equal to, let's write it this way, this is going to be the limit, I'll do all of it in blue actually. This is going to be equal to the limit as, let's introduce a new variable, B approaches infinity of the integral from six to infinity of one over N squared DX, which is equal to the limit as B approaches infinity of, let's see, the antiderivative of this is negative N to the negative one, and we're going to evaluate that at B and at six, so this is equal to the limit as B approaches infinity of negative one over B, negative one over B minus negative one over six, so plus one over six. Well as B approaches infinity this term right over here goes to zero, so this is just going to be equal to one sixth, and we can use that exact same logic to evaluate this integral, so let's do that. Actually, I'll do it right over here. So the improper integral, I should say, from, and here we're going to evaluate not from six but from five, that's our K. Five to infinity of one over N squared DX. Well, the only difference here between these two integrals is this bottom bound. This was one over six, and by the same arugemnts, this one is going to end up being one over five. And so now we can put everything into this compound inequality. We're going to have one point four six four. One point one six four plus, and I'm being a little bit approximate here, it's really this S sub K, which we have the exact value for in our calculator, so it's one point four six four plus one sixth, plus one sixth, is less than or equal to, less than or equal to, is less than or equal to our sum, which we care about, which is less than or equal to one point four six four plus one fifth, plus one fifth, and so this one is fairly straight forward to calculate. This is going to be plus point two, so this is going to be less than or equal to one point two six four, so sum is going to be less than or equal to that, and on this side if we took our original expression, and then I added one sixth, one sixth, I get one point...I get one...let's see, what did I? Oh, this is supposed to be one... I wrote this one wrong. O.K., so this is going to be one point sixth three oh. Let me just write one point... So it's going to be one point six three oh, and on this side, my brain was malfunctioning. This should be one point one, I took away two instead of adding two, one point six six four, one point six six four, and so just like that you see, O.K., this thing must be like one point six three something. It could be one point four something. It could be one point five something. It could be one point six six, but this is already giving us a lot of precision, if we only cared about things to the 10ths place, this kind of, we're done. We know it's going to be one point six something, and if we go a little bit further this is already giving us a lot of precision, and you can imagine if you added a few more terms in the partial sum, you would get even more precision. So hopefully you found this to be kind of neat exercise. Not only did we get a good bit of precision, a good bit of precision with just taking a partial sum and the help of these integrals, but we got a good bound on it. We know that the sum that we're converging to is between these two values.