In this video, I go over the Integral Test, which states that a series converges only if its corresponding improper integral also converges. Likewise, if the improper integral diverges, then the se...
Transcript
the integral test and this integral test states suppose f is a continuous positive decreasing function on the interval from one into up to infinity and let a n equals to f of n and this f of n is the same thing as i've done over here notice here this is for this case the a n would be one over square root yeah the area is just going to be one over square root one in other words f of n here so y equals f x there one over square root x but that was just a specific one i'll go over the general proof in later in this video at the end so yes if we're given this then the series of right here from summation of n equals 1 up to infinity of a n is convergent if and only if the improper integral and I'll write this down right here, 1 to infinity of f x d x. Yeah, so if this is convergent, let's just delete. The empty space. Yeah, so, yeah, so this is, the series is convergent, if and only if the improper integral right here is also convergent, or is convergent, then the other one would be. So, in other words, if what we have is, if this is, if the limit of, or the integral from 1 to infinity of FX, DX is convergence, this is the I part of it. then the series, summation from n equals 1 to infinity, if an is also convergent. But then if this isn't convergent, so I'll just write this down again for completeness if, integral from 1 to infinity of F of X, DX. Yeah, so if this is divergent, so if that integral underneath or the area underneath the function is divergent, then what we have is the series from n equals 1 to infinity of AN is divergent. So yes, exactly as those two examples I did above. And note that when we use the integral test, it is not necessary to start the series or the integral at n equals 1. For instance, in testing the series, if we want to test this one right here, from n equals 1 to infinity of the summation 1 over n minus 3, squared, actually instead of 1, we have 4. So if we're starting from n equals 4 like that, we use instead of n equals 1, we use n equals 2, yeah, we use n equals to 4. So we go from the integral from 4 to infinity, not 8 infinity like that, of 1 over x minus 3 squared d x. So yes, it doesn't necessarily need to be n equals 1. And also, it is not necessary that the function F be always decreasing. What is important is that F is ultimate or be ultimate decreasing. That is decreasing for X larger than some number n. So yes, as it goes to infinity, as long as there's some number that greater than it is always decreasing. Then the series, summation n equals capital n, up to infinity of A.N is convergent. This is also convergent when we start from n equals 1 to infinity of a n is convergent by note 4 in my previous video on infinite series. So, yes, doesn't need to always be decreasing as long as eventually it starts decreasing greater than a number. And this is from note 4 in my infinite series video, again from Steam it, there is a screenshot from my calculus book. So a finite number of terms doesn't affect the convergence or divergence of a series. For instance, suppose that we're able to show that the series, right here, summation from n equals 4, as opposed to n equals 1, to infinity of n over nq plus 1 is convergent. So if that's convergent, since we have here, we have like this, so if you're looking at n equals 1, to infinity of n over n q plus 1, then we could just add the first three terms. So 1, 1 over 2, plus 2 over 9, plus 3 over 28, plus this one, this one's convergent. So because this one's convergent, This is just summation of just finite terms. So it follows that the entire series, yeah, from n equals 1 to infinity of n over n q plus 1 is convergent. Similarly, if it is known that the series right here, from the summation from n equals capital n plus 1 to infinity of a n converges in the full series starting from n equals 1 to infinity of the summation a n equals 2 from n equals 1 up to capital n of a n plus n equals capital n plus 1 Yeah, to infinity of AN. So yes, these are just finite terms. This one, you go into infinity, as long as that converges, then everything else converges, because these are just finite terms. So that's also converging. So, yes, that follows exactly as I was stating here. It doesn't need to be always decreasing as long as eventually, it's decreasing. Alright, so now let's look at example 1, just to test out this integral test. So test the following series for convergence or divergence, and we have here from n equals 1 to infinity of the summation 1 over n squared plus 1. So this is 1 over n squared plus 1. As you can see here, as you increase the value of n, this becomes what you're dividing by, or this terms become dividing by a larger number. In other words, it starts to decrease as you increase n, so it's decreasing. and it's also, here's always positive. So it's always positive, and it's always decreasing it in, and it is continuous, because there's no discontinuities here. You can't make the bottom zero. So the function F of X equals 1 over x squared plus 1. So it's continuous, it's always positive, and it's decreasing on the, on from an interval from 1 to infinity. So we use the integral test, and the integral test is, again, just take the integral 1 over, some integral from 1 to infinity. of 1 over x squared plus 1 d x this equals 2 this equals 2 we so we could take the limit as t approaches infinity of integral from 1 to t of and using improper integral techniques 1 over x squared plus 1 d x like that and now to solve this integral i have to recall from my earlier video on the derivative of inverse 10 of x or 10 negative 1 of x or arc tan or arctangent or whatnot however you want to call it from my earlier video is linked to it one of my very earlier ones so and then if you go through that proof it goes over it and it shows that d over d x of inverse tan of x is equal to 1 over x squared plus 1 so in other words we can take the anti-derivative or the integral and to get back to this one right here so that's that's what we can simplify that so thus So thus we have, so thus this part right here, limit as T approaches infinity of, from 1 to T, of 1 over x squared plus 1, DX, this equals to, yeah, this equals to the limit, as T approaches infinity, and then we can just box this inside, and that's going to be our... this is going to be the antiderivative, is going to be tan, inverse tan of x, like that. Because you take the derivative, it gets over that, like that, and this is going to be from 1 to T. So, this equals to, we're going to have a, equals to a limit, as T approaches infinity of, of, now we're going to have a brackets, Inverse 10 of t, minus, put a bracket like that, minus 10 of 1 like that. And now to solve these two parts right here, and this one, as T approaches infinity of that one. What I want to first calculate is, or solve is this one first right there. So let's put a line here, recall, or our note, basically using exact trick ratios, but here we can just go easily and draw a square. So we draw a square with sides 1, 1, 1, 1, like that, and I'm going to split this in half. So the angle here is going to be 90 degrees or perinicular, or in radiance, pi over 2. So what I'm going to do right now is draw this or cut this in half. so that the angles divided by 2. So in other words, now we have, this is going to be 45 degrees, or pi over 2 divided by 2, this can be pi over 4. And it's going to be pi over 4 like that as well. So now, this is the right angle. So what we can do is just take the tangent, recall tangent of this angle, pi over 4, is defined as opposite over hypod, or I'm opposite over adjacent. Adjacent. So if it's like that, the opposite is just 1, divided by 1 is the adjacent. So we have 1 over 1, that equals to 1, like that. So that's, yeah, that's 1 over 1 equals 1. So now, but we want to know the inverse 10, number inverse, all we do is switch the x and y, or switch this with that one. So in other words, we get the inverse 10 of 1 is just equal to pi over 4. So tan pi or 4 equals 1, the inverse is going to be, you switch those around. So the inverse of, yeah, inverse tan of 1 equals to pi over 4. So there is that this value, this is just pi over 4, like that. And now this one here, we're going to take, we have to take a limit there, so that's going to be a bit interesting there, so I'm going to draw this again. Yes, and here I'm going to write also recall, this one, also recall, trigonometry. Very interesting stuff. So we'll have a curve like this, Y, and there's the X, there's the Y, and what I'm going to do is I'm going to graph, recall the graph of tangent, it looks like this, it goes downwards like that, and for the inverse, they're usually defined from pi over 2 to negative piper 2. So here, this curve right here is our y equals to 10, I'll also over 10 of x. I'll put it over here. So this is y equals to 10 of x, like that. So we have that, and this line right here, asymptope line all the way going down like that, and this asymptope line is at pi over 2, and then this asymptope line across here is at negative pi over 2. Like that. And here, actually just to align with this T there, I'm going to write this as T. Change a variable from X to T. Yeah, so now to draw the inverse, I was going to write inverse, or better yet, write it in red. So inverse is effectively just switching the, is switching the Y and X. Or the coordinates around, so switching X and Y to y and x so in other words here so if this one's here if it's approaching if or y and t in this case actually t and y and y and t so if as t approaches pi over two here this goes to infinity so then the reverse is going to happen so that as t approaches infinity then the y goes to pi over two so we could draw the asymptote lines like this It's going to be across like that, and then likewise for here. So as T approaches negative pi over 2, then Y goes to infinity, but instead it's going to be opposite, so then as Y approaches negative pi over 2, T goes to infinity, or negative infinity. Because this is going negative infinity like that, so we'll just draw asymptote like this, this is like that, and this is going to be an all right. this down, this height is pi over two, this one's going to be negative pi over two, so then this is going to be inverse in every way. So it's going to go, yes, it's going to start here, it's going to just touch it, and then go all the way across like that. And likewise, here, it doesn't actually overlap, but instead it goes like this. Let's try this a bit better. Oops, draw this a bit better. Yeah, like that, and it goes all the way. Yeah, so draw it better. And this is the inverse, so y equals to 10 inverse of T, like that. Yeah, so here, as T is going to infinity, over here, as you can see, over, that else write this there, T is approaching infinity, then Y is approaching pi over two, so the inverse of this one there. So we could, yeah, this is going to be reaching pi over two. yes so let's put it down here so this one right here is uh yeah this is approaching pi over two as you plug in that t goes to infinity there this one has no t there so this just is just pi or four so what we end up having is so thus we have yes so that's the integral so the integral so the integral right here of one over uh infinity it's just a from beginning so one over no integral from one to infinity of one over x squared plus one So, write it all down here. So integral from 1 to infinity of 1 over x squared plus 1 d x. This equals to pi over 2, yes, you know, finally, so pi over 2 minus pi over 4. But pi over 2, it just times the top and bottom by 2 over 2. This just equals to, well, 2 pi over 4 minus pi. That's just 2 pi minus pi. This equals to pi over 4, like that. There we have it. Pi over four. Fascinating stuff. Yes, so thus what we have is the integral from 1 to infinity of 1 over x squared plus 1. DX is a convergent integral, and so by the integral test the series, as well as, yeah, because the integral is convergent, thus the series is as well, the series of 1 over n squared plus 1 is also convergent. So it's fascinating, fascinating stuff.