In this video, I go over an introduction into estimating sums of series by examining whether their corresponding improper integrals converge or diverge, thus providing the basis for the Integral Te...
Transcript
Introduction to the Integral Test and Estimate of Sums In general, it is difficult to find the sum of exact, yeah, to find the exact sum of a series. Remember, series is just a sum of a sequence. If, yeah, we were able to accomplish this for geometric series, and the series the series right here is summation of a summation of 1 divided by n times n plus 1 in my earlier videos because in each of those cases we could find a simple formula for the nth partial sum s n but usually it isn't easy to compute limit as an approaches infinity of s of n. Usually we can't compute the limit as the partial sum goes to infinity or even find this partial sum. Therefore, in the next few sections, we develop several tests that enable us to determine whether a series is convergent or divergent without explicitly finding its sum. In some cases, however, our methods will enable us to find good estimates of the sum. Our first test involves improper integrals. We begin by investigating the series whose terms are the reciprocals, or i.e. multiplicative inverse, for example, 1 over 2. is a reciprocal of 2 since 1 over 2 times 2, so you multiplied, or the multiplicative inverse, you multiplied by its inverse, equals 1, like that. So 1 over 2, so for example, 1 over 3 would be the reciprocal of 3. So we're asked, we begin by investigating this series whose terms are the reciprocals of the squares of the positive integers. So yes, that's a wording as a mouthful for my calculus book. But anyway, so what I mean by that is, let's take the summation from n equals to 1. up to infinity, and this is going to be, well, the reciprocal of the positive integers, but the square of them. So we're going to go 1 over n squared. Yeah, so that's the positive integers. So, in other words, it's going to be summation of 1 over 1 squared, plus 1 over 2 squared, plus 1 over 3 squared. So it's reciprocal of the positive integers, of the squares of the positive integers, hence the squared over there, plus 1 over 4 squared, plus 1 over 5 squared plus dot dot dot and go on Let's see, so there's no simple formula for the sum S-N or the partial sum of the first N terms, but the computer -generated table of values below suggest that the partial sums are approaching number near 1.64 as N approaches infinity, and so look, yeah, and so it looks as if the series is convergent, and here's the table, and I must note the Excel file, and the link below will be used throughout this document, so make sure to download and play around with it, so I have it in there. So here is the table, and here is all the way to... have 5,000 terms there. So we have term is 1 over n squared, so I just put this inside, divide it out, so 1 over 1 over 1 square is just 1. And then 1 over 2 squared is going to be 1 over 4, so 0.25, and then this is the summation of them. So 1 plus 1.25, or I mean 1 plus 0.25 is equal to 1 .211, and then 1.25 plus 1,111 there. We get this over here, 1.3611. And so on, they go all the way to 5,000 terms, we get this 1 .6. 4 figure. So yes, it appears to be approaching, because the number, the first term is 1, and then it goes all the way here, the summation of all of them, you just get to 1.64. So it does look like it is converging, but that's not fully proof. But what we can do is we can confirm this impression with a geometric argument. So the figure below shows that the curve y equals 1 over x squared and rectangles that lie below the curve, and I'll actually draw that right now. So let's say we have like this, this is the x, that's the y, and let's say we have our function y equals 1 over x squared, so let's say it looks like this, this is going to be y equals 1 over x squared, so as x gets larger, this is going to be divided by large number, so it's approaching zero like that. Or actually, let's make this a bit more steep, just like this, all right. and now what I'll do is I'll just write this 0 there. Let's say we have our first term, or our first X coordinate 1, then we have 2, then we have 3, then we have 4, then we have 5 here, and so on. And here I quickly just change a curve again, just to make it a bit more presentable, because I'll be drawing some rectangles. So what we have here is let's draw a rectangle from this one coordinate there all the way up top here, and it goes like that. So now this distance from here to there, that's just 1. So in other words, we're going to have a rectangle here with height. So this height right here is going to be, well, 1 is going to be the y. So 1 over 1 square. In other words, this area equals 2 is going to be 1 square times 1. And yeah, that all is always going to be times 1. So this is going to be 1 over 1 squared. That just equals to 1 for this first 1. Like that. But likewise, I'm just going to erase this. It's pretty straightforward. So likewise, from this rectangle here, what we end up having is, then this goes like that. This area here, this is area equals to, well, 1 over this new y is going to be 1 over 2 squared. And then times by 1, like that, so it's going to be 1 over 2 squared. And likewise over here, we'll have another rectangle like this, and this one's going to be area. equals to 1 over 3 squared. And then we have another one here. Notice how it's always below the curve. So here, and this is our series that we were looking at. 1 over 3 squared. I mean 4 squared, like that. And then finally here, we got just another 5th one. This area here is, I'll just put it here, area equals to 1 over 5 squared. So yes, here I've written up here. So the base of each rectangle is an interval of length 1, as I drove over there. The height is equal to the value of the function, 1 over x squared. So here, like that, at the right endpoint of the intervals. For intervals from 0 to 1, we look at the right end point, from 1 to 2, we look at the right end point 2, and so on. So the sum of the area of the rectangles here, what these end up becoming are, if we just sum up all these, those rectangles, we have 1 over 1 squared, plus 1 over 2 squared. plus 1 over 3 squared plus 1 over 4 squared, plus 1 over 5 squared, and it keeps going on and on. You can keep going on to infinity, so we get equals to 2, like this, equals to the summation from n equals to 1, up to infinity of 1 over n squared. All right, now if we exclude the first rectangle, since fx goes to infinity as x approaches zero, so here, if we just exclude this one right here, just so we don't need to deal with this infinity function going all the way infinity as x approaches zero, because 1 over 0 is infinites, just so we don't need to deal with it. So the total area of the remaining rectangles is smaller than the area under the curve y equals 1 over x squared, 4x is greater than... or greater than or equal to 1. So here, if we look at these rectangles, sum those are, all those are, those are, those are going to be less than this, the area underneath this function, because you can see that there's a remainder every single area here, this gets smaller and smaller as this goes further and further. And, yeah, so it's smaller than that, which is the value of the, yeah, so that area, of the underneath this curve, which is going to be greater than the rectangles area, which is the value of the integral from 1 to infinity of 1 over x squared. dX so again one instead of the zero so you know don't need to deal with the second infinite so recall in my earlier video on improper integrals that this improper integral is convergent and has a value of one so it's from a steam it right here the steam at notes in my earlier video several years ago this is improper integrals type one and infinite intervals so yes when we have an infinity sign it's going to be an improper one we're going to have to do what I showed in that earlier video here and I just took a screenshot of the calculus notes is it a bit neater than mine, and more compact. So here's the function why goes 1 over x squared. So if you take the integral, or the area underneath this from 1 to t, is going to be, well, 1 over x squared, take integral, you get this, 1 minus 1 over t, that's the area underneath it up to t, and then we take the limit as t goes to infinity, and it's going to be 1 minus 1 over infinity goes to 0 there, so equals 1. So that entire area is going to be going towards 1. and hence is convergent and it has the value of 1. So the area of the shaded region approaches 1 as T approaches infinity, see a figure there. So we say that the area of the infinite region S is equal to 1, and we write over here that the integral from 1 to infinity of 1 over x squared dx equals to the limit as T approaches infinity of 1 to 1 to x, from 1 to the integral of 1 of x square d x equals to 1. So the picture shows that all the partial sums are less than. and just put in brackets, including the first rectangle, so that's going to be, well, the first rectangles will be 1 over 1 squared, and then we're going to take the, it's going to be less than this, remember this, then it's going to be less than the area underneath the curve, which is 1. So the area underneath this curve here, so we're going to add this up right here, and then area underneath the curve. So the first area plus, and then it's going to be less than this right here, integral from 1 to infinity of 1 over x squared, d x, and this one right here is equal to 1. So this, or I just put this like this, this equals to 1. Then add these up, that's just going to be 1 plus 1, equals to 2. So the total area of those partial sums has to be less than 2. So thus the partial sums are bounded. So yes, they don't go to infinity, they go to below a certain number. So we also know that the partial sums are increasing because all the terms are positive. So you're always, yes, so you have this 1 over n squared right here. This number is always positive, so you're always adding to it. So it's always gradually increasing. Because the terms are positive. Therefore, the partial sums converge, and this is by the monotonic sequence theorem from my earlier video on infinite sequences and series. So the series is convergent. Thus, the sum of the series, and the sum of series, or the limit of the partial sums, same thing, is also less than two. So what we could do is write. Like this, infinity, n equals to 1, and I'll recap on monotonic sequence theorem soon. So 1 over n squared, like this. This equals 2, 1 over 1 squared, plus 1 over 2 squared, plus 1 over 3 squared, plus 1 over 4 squared, plus 1 over 5 squared, plus dot, dot, dot, this is all less than 2. So we have concluded by using just integrals there. That's less than 2 there. Now recall the monotonic sequence theorem from my earlier video on infinite sequences, and here's a statement article from there of the video notes, as well as here's the screenshots from the calculus for more compact, and so a sequence here, just the monotonic sequences, here's every bounded monotonic sequence is convergent, and this is the definition, just to recap, sequence AN is called increasing if ANN is always less than A & plus 1, 4N is greater than 1. that is A1 is less than A2, less than A3, and so on, it is called decreasing if it's the opposite AN is greater than A, and plus 1 for all N greater than one. It is called monotonic if it is either increasing or decreasing. So yes, it's increasing, it's monotonic there. And a sequence A-N is bounded above if there is a number M such that A-N is less than M. for all A.N greater than 1. And since the function is going to be less than 2, it's going to be obviously bounded. So you can't keep going higher and higher, and it's going to be bounded above, since A.N is less than M for all N is greater than 1. and likewise it is bounded below if there's a number m such that the reverse a n is greater than it's lowercase m for all n is greater than one if it is bounded above and below if it is bounded above and below then then a bounded sequence so yes if it can't go below at a certain point or a higher than a certain point and and the one we're dealing with well it can't go below Yeah, it can't go below 1, because it's always increasing, or it's always decreasing its value, but it's positive, so increasing the partial sum. So yes, it can't go below this one here, because you're always adding more and more to it, and so on. So going further, and then, so, yeah, there's monotonic sequence. Every bounded monotonic sequence is convergent. Essentially, just do a recap on it more, if a sequence is always increasing or decreasing, hence monotonic, and is bounded above or below, then it will approach a number or limit L, which is less than capital M, or L is greater than M, or greater than or equal to M if the sequence is decreasing. For example, here, if it's going to be less than this number M, there's going to be some sort of limit, it could be equal to M there, if it's bounded below. converges all the way to it, so there's a limit of that. And that's pretty much a monotonic sequence, because this is always increasing, and it's bounded above, below, and so on. Note that the exact sum of this series was found by the Swiss mathematician Leinard Euler, 1707, 1783, that so he was born and died, to be pi squared over six. But the proof of this fact is quite difficult, so I will have to prove it in a later. video and it's another problem plus video, so I'll do that much later on. All right, so now let's take a look or let's look at the following series, and it's going to be a slightly different one. So instead of doing the squares, we do the square roots. So we have one over, I mean, it's sum from n equals to 1 to infinity of 1 over square root n. So this equals to 1 over square root n, plus 1 over square root 2, plus 1 over square root 3, plus 1 over square root 4, plus 1 over square root 5, plus dot, dot, dot, and it just keeps going and going. Yeah, so the table of values of SN, of the partial sums of these suggest that the partial sums aren't approaching a finite number, so we suspect that the given series may be divergent, and that's this table, it's again from that same Excel sheet, and here's the terms from one all the way to 5 ,000 there, and here's the term 1 over square root, and or end of power of 1, 1 half, same thing. So 1 over square root 1 is 1, and so on here is 0.707 for squarely 2. And then here's this partial sums where we add these up. So when we add these up, 1 plus 0.7, we get 1.707. And then we go to 5,000. Notice this was 139. So the 1,000th term, thousands partial sum is 61, 4 is 43 for 500. So it just keeps going bigger and bigger. Again, we use a pitch. for confirmation. So yeah, so this case looks like it's diverting. So the figure below, yeah, figure out of the text, that shows the curve y equals 1 over square of 2, or x power of 1 half. But this time we use rectangles whose tops lie above the curve. So here I'll draw this one out. So what we end up having is like this. So there's the x. there's the y and this curve is going to be a yeah it's not going to be as steep as before it's going to be more like this let's draw a bit better it's going to be something like this all right so let's say we have that this is our y equals to 1 over square root x curve like that let's draw our coordinates here or the x xx so there's zero let's say we have one is going to be somewhere here one then we have two, then we have three, and then we have four, and here I just shifted everything out of the fifth one there, and then made the curve a bit more to the left so that we can do better drawings. Yeah, so we're going to draw the rectangles above the curve, so in other words, if we start from here, we're going to use the left end point as opposed to the right one, so the rectangles can go across this way, like that. so that this area right now, this height is going to be, well, 1 over square root 1 is the height, and then the width is just 1. They're all going to be separated by 1. So this area inside, is equal to, well, 1 over the height times width, and yeah, that's just times 1. So 1 over square root 1 times 1, which is, we just can be that. And then likewise, for this side here, we're going to get, you can use the left end point, so that the rectangle is always above the curve. So now we have the area here. This is using the left end point there, too. This equals to 1 over square root 2. And likewise, we go across like that, and then we use this one here. This one is also area. Now we use the third one equals 1 over square root 3, and then finally here we'll go like that. It's going to look something like this, where this area is equal to 1 over square root 4, again, times 1, that's just unit length, so it just remains the same. Yes, and so on, this one's going to be the same thing across over there, and again, assuming this is always decreasing like that. Alright, so the base of each rectangle is an interval of 1 or the width there. The height is equal to the value of the function 1 over 1. divided by square root x or x power of one half, at the left end point of the interval. So that's the value of this height there. So the sum of all the rectangles is, well, we just sum up all these rectangles like that. So it's going to be 1 over square root 1, plus 1 over square root 2, plus 1 over square root 3, plus 1 over square root 4, and so on. So 1 over square root 5. plus dot, dot, dot, this equals to, right, this has a summation, n equals to 1 up to infinity, and this is going to be 1 over square root n. Yeah, so now this total area is greater than the area under the curve, Y equals 1 over square root x, 4x is greater than 1, which is equal to the integral, so integral below this curve right here, so this is always above it, so it's going to be greater than the integral. I'll write it down like this, so yes, it's going to be greater than this integral from 1 to infinity of 1 over square root x d x, like that. Yes, but we know from my earlier video that this improper integral is divergent, and this is from my early videos, so yeah, hopefully you've been paying attention to my earlier video from years back. So this improper integral is example 4, here's from Steamit as well, the video notes, and I've screenshot it. the calculus version of the notes just it's more compact so for what values of p is the integral right here from 1 to infinity of 1 over x p d x is a convergent so for what values of that is convergent and then here the solution shows that or from previous one is show that p equals 1 is going to be divergent so 1 over x and if it's if p is not equal to 1 then you have then there's the proof over here so if Over here it has if p is greater than 1, then the integral converges, but if p is less than 1, then over here, the integral diverges, and it summarizes here as integral from 1 to infinity of 1 over x to power of p, dX is convergent, if p is greater than 1, and divergent if p is less than or equal to 1. In other words, area under the curve is infinite, because here we have, this is 1⁄2, which is, again, less than. 1, so that's the p, and that's less than 1. So, in other words, the area under the curve is infinite. So the sum of the series must be infinite, because the area under the curve is infinite. So if this area under the curve is infinite, but these rectangles are more than that infinity, so it's also infinity. So the series of the, yes, the sum of the series must be infinite, that is the series is divergent. The same sort of geometric reasoning that we use for these two series can be used to prove the following test. The proof is given at the end of this section, and it's called the integral test.