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

Video transcript

in the last video we saw that if we had an infinite series where each term is a function of N and the function itself is a continuous positive decreasing function over the intervals that we care about and we assume that this series converges that we can estimate that series we can just estimate that value that it converges to using a finite number of terms and some integrals on the way that we establish these integrals is because when we split up the sum into a finite sum and then another infinite series we were able 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 that allowed us not only to 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 and so let's see how good these are by applying it to a particular infinite series so let's say let's say that the infinite series we want to apply to let me do it in the yellow color so it's going to be let's say we go from N equals 1 to infinity of 1 over N Square and let's say we didn't know how to find the exact value here but we want to estimate it and let's say we want to estimate it using the first let's say we want to estimate it using the first five terms so we'll do s sub 5 is equal to the sum the sum of N equals 1 to N sorry it equals 1 to 5 of 1 over N squared and let's see this is going to be this is going to be equal to let's get a calculator out this is going to be 1 over 1 squared which is this 1 plus 1 over 2 squared which is 1/4 plus 1 over 3 squared which is 1/9 plus 1 over 4 squared which is 1/16 1/16 plus 1 over 5 squared which is going to be 125th 125th and we get this value right over here so I'll just say that this is I'll say it's approximately 1.4 say for approximately one point four six four and then if we let's 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 ten terms or twenty terms and we would have a worse estimate if we did K equals three but I just pick a equals five because it seems reasonable something that we could compute I actually didn't need a calculator to do that it could've done it by hand it would have just taken more time but 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 that's going to be equal to let's write it this way this is going to be the limit we do all of it in blue actually this is going to be equal to the limit as the limit as let's introduce a new variable B approaches infinity from of the integral from six to infinity of one over N squared DX which is equal to the limit as B approaches infinity of of let's see the antiderivative of this is negative negative n to the negative one and we're going to evaluate that at B and it's six so this is equal to the limit as B approaches infinity of negative 1 over B negative 1 over B minus negative 1 over 6 so plus 1 over 6 well as B approaches infinity this term right over here goes to 0 so this is just going to be equal to 1/6 and we can use that exact same logic to evaluate this integral so let's do that actually so I'll do it right over here so the definite the improper integral I should say from and here we're going to evaluate not from 6 but from 5 that's our k5 to infinity of 1 over N squared DX well the only difference here between these two integrals is this bottom bound this was 1 over 6 and by the same arguments this one's going to end up being 1 over this one's going to be one over five and so now we can put everything into this compound equality inequality we're going to have one point four six four one point four six four plus this is 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 six plus one six 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 for one point four six four plus 1/5 plus 1/5 and so this one is fairly straightforward to calculate this is going to be plus point two so it's is going to be less than or equal to let's see one point two six four so so I'm just going to be less than or equal to that and on this side if we took if we took our original expression and then I added 1/6 1/6 I get one point I get one let's see what did I oh this is suppose to be one and I wrote this one wrong ok so this is going to be one point six three oh let me just write one point so it's going to be one point six three oh and on this side I know it my brain was malfunctioning this should be one point one I took took away two instead of adding to one point six six for one point six six four and so just like that you see okay this thing must be like one point six you know it could be one point six three something could be one point six four something could be one point six five some think we won one point six six but this is already giving us a lot of precision if we only cared about things to the tenths place this kind of we're done we know it's going to be one point six something and if we will go a little bit further this is already giving us a lot of precision and you could 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 an eat 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