Original subtitles

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In this video we're talking about how to use the integral test to say whether or not a series converges

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and the integral test is one of our many convergence tests.

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In this particular problem we've been given this series here one divided by n square plus one we're

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taking the sum of that series from end equals one to infinity.

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And what we want to do is use the integral test to say whether or not this series converges or diverges

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so the integral test tells us that if a sub that's our series here one over and square plus one is positive

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decreasing and continuous.

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So these three points here are extremely important if our series isn't positive and or is not decreasing

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and or is not continuous we can't use the integral test at all to determine convergence or divergence.

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We'll have to use a different test or maybe we won't be able to determine convergence at all.

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So we have to first make sure that this series is positive decreasing and continuous.

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If it is all three of these things then we can move on with the integral test and then the integral

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test tells us basically that the behavior of this series and the behavior of this integral the integral

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of f of x from 1 to infinity is the same meaning that either both of these things converge or they both

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diverge.

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So in other words it gives us something else we can evaluate to determine convergence.

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So we can find the integral from one to infinity of this function f of x and we just set a 7 and f of

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x equal to one another so we use this value here one over end squared plus one.

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We just switch out the variable and for the variable x and then we would take this integral integral

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from one to infinity of 1 divided by x squared plus 1.

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We'd evaluate that integral and then if that integral converges we know that the original series converges

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if that integral diverges.

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We know that the original series diverges because the integral test tells us that these two things will

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always have the same behavior.

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As long as the series meets these first three criteria and therefore allows us to use the interval test

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in the first place.

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So let's tackle these first three criteria.

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First we have to show that the series is positive decreasing and continuous.

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So we'll start with positive and oftentimes the easiest way to figure these three out is just to find

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the first couple of terms and look at the trend that you see between them.

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So what this tells you here this and equals one on the sun.

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This tells you that the series starts at end equals 1.

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So I'm going to plug one in for n then I'm going to plug into them 3 and 4 then 5.

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And those are the first few terms of this series.

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So if I plug in and equals one I'm going to get one squared which is one.

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One plus one is two one divided by two is one half so the first term in the series is one half the next

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term in the series is going to be the termit and equal to Pelion n equals 2 I get two square which is

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4 4 plus 1 is 5 1 divided by 5 is one fifth.

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Then I'm going to plug in and equals three.

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I'm going to get nine plus one is 10.

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So I'm going to have one tenth then I'll play again and equals four.

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I'm going to get 16 plus 1 or 17.

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So I'm going to get one over 17 so you can see the pattern here these are the first four terms of the

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series and I couldn't keep going.

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Now I want to look at these terms and say Can I use these terms to say whether or not the series is

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positive.

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And in fact I can I can see that the terms are getting smaller.

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One fifth is smaller than one half.

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One tenth is smaller than one fifth 117 is smaller than one tenth.

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But no matter how small they get the terms are always going to be positive.

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There's no value of.

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And I can plug in because my end values are always going to be positive.

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They're going to start at positive 1 and count up forever positive two positive three positive four.

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So I'm always going to plug in a positive value and there's no value I can plug in for n that's going

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to make this term negative therefore I can say that the terms in my series are always going to be positive

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and therefore that the series itself will always be positive then decreasing.

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So if I look at my series here I can see that one half is greater than one fifth one fifth is greater

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than one tenth one tenth is greater than 1 17th.

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The numerator is always going to stay one.

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The denominator is going to get larger and larger and larger which means the fraction as a whole gets

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smaller and smaller and smaller.

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So every term is always going to be smaller than or less than the previous term.

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And when each term is less than the previous term then I can say that my series is decreasing so that

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one's taking care of as well.

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Now what about continuous.

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Well basically I just want to show that there are no discontinuities in the domain of this series and

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I know that's true because since my series starts it equals one and I count up and equals positive 1

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to posit three positive for on into infinity.

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I'm always just going to have a positive number in the denominator of each term.

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I'm never going to be able to plug in a value for n that's going to make this denominator zero and therefore

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result in a discontinuity of the function there's no way I can make this function discontinuous so I

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can say that the function is in fact continuous therefore I can say A7 the series is positive decreasing

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and continuous so I know I can now use the integral test.

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So use the integral test what I want to do is I want to say a Subban is equal to f of X..

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So in other words this value here my series that's a seben that's this value right here.

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So then I want to set a sub an equal to f of x and I just want to put it in here into this integral

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for f objects.

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The only thing I need to do because my integral here is in terms of x variables.

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I just want to go ahead and change the end variable to an x variable and then evaluate this integral

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so I'm going to have the integral from 1 to infinity and then ever X is going to be dysfunction here.

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One over.

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But now instead of and I want to say X so I'm going to say X squared plus 1 D x.

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Now we need to evaluate this integral and figure out whether or not this integral converges or diverges

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if it converges in the original series will also converge if it diverges than the original series will

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also diverged.

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So to take the integral what we need to realize is that this function here one over x squared plus 1

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is a special type of function and this just happens to be this particular problem.

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But the integral of this quotient is arctan or Tanev the negative one of x and this function actually

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comes up a lot in calculus so it's helpful to know that the specific value 1 divided by x squared plus

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1 the integral of that function is arctan of x.

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So we're going to have arctan of X evaluated over the interval 1 to infinity.

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Now I'm going to be technical about this what I want to do is change this to the limit as B goes to

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infinity arctan or 10 and negative 1 of X evaluated over the interval one to be.

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And so then I would say the limit as B goes to infinity and I actually could have done that before I

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took the integral but I've got the limit as Bigos to infinity evaluating over the interval.

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I plug in my upper limit of integration first which is B.

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So I'm going to get arctan of B and then minus whatever I get when I plug in my lower limit of integration

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once I'm going to get minus arctan of one.

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Now you can plug these values arctan of infinity and arctan of one into your calculator or you can do

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the math and use the unit circle and then we'll just take arctan of one as an example here.

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If we say 10 to the negative one of 1 is equal to x and then we take a tangent of both sides.

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We're going to take tangent of the left side here and then tangent of x we get tangent and arctan to

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cancel and we get one is equal to tangent of x we remember that tangent is sign over cosign.

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So what we get is sign of X over cosign of x is equal to 1 and the only place where this happens on

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the unit circle member on a unit circle sign represents the y value of each coordinate point cosen represents

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the x value each coordinate point so really what we're looking for is why over x is equal to one or

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the point where y is equal to x.

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The only point where that happens on the unit circle is the angle PI over 4.

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So there's arctan of 1 is PI over 4.

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You could also do that with arctan of infinity or taking the limit is Bigos to infinity so we're kind

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of plugging infinity in for be here.

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And if we did the same Matthew what we get is y over x is equal to infinity instead of Y over x is equal

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to 1.

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The closest we ever get on the unit circle to infinity the largest value we ever obtain is that PI over

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two.

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So we're going to say is that arctan of B is PI over 2.

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So we end up with PI over to minus PI over 4.

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We can multiply both the numerator and denominator of this first fraction by two to get a common denominator

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so we end up with two pi over four minus PI over four to PI minus pi is just pi.

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And then we have four in the denominator.

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So our answer then is PI over 4.

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What's important here is not the exact value of the result.

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What's important is whether or not you get a real number answer here or an infinite answer.

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So the fact that we got a real number pi over 4 is a real number.

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It's a constant.

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The fact that we got a real number means that this original integral here converges.

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So a real number answer indicates that this converges.

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If we had found instead a result of infinity or negative infinity that would indicate that this original

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integral was divergent but because the value of this integral was a real number that means the integral

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converges and the integral test tells us that when this integral converges the original series also

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converges so we can say that this original series is convergent.

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Again if we had found an infinite value here that would mean that the integral was divergent and we

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would say that the original integral was also divergent.

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So the last thing I'd say is the key to knowing whether or not you should use the integral test is to

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look at this original series and see whether or not it's easy to integrate because remember you're just

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going to take this function and plug it into an integral.

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Yes we change the variable from end to X but we just evaluated this series as an integral.

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So if this original series is easy to integrate and you know you can integrate it then you might be

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able to find the value of this integral.

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So the integral test might be a good choice.

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Sometimes you're going to get a series that's impossible to integrate if you imagine the series in an

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integral and you think to yourself how would I ever take the integral of this series the integral test

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might not be the test you're looking for.

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But if this looks easy to integrate then you might want to try the integral test and then you should

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go immediately to this positive decreasing continuous checklist because if this series is positive decreasing

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and continuous and that's when you know you'll be able to use the integral test.

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And again the result of the integral.

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All that matters is whether or not it's a real number answer in which case this integral converges and

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so does the series or it's an infinite answer in which case the integral diverges and therefore so does

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the series.