Confirm that the integral test applies and then use the integral test to determine convergence of the series. (A) ∑ n = 1 ∞ a r c tan n n 2 + 1 \sum_{n=1}^{\infty}\frac{arc\tan n}{n^2+1} (B) ∑ n = 1 ∞ 1 3 n + 2 \sum_{n=1}^{\infty}\frac{1}{3n+2}

this problem, we're asked to confirm that the integral test applies, and then use the integral test to determine the convergence of the following series. Now, we're going to go ahead and start with example a, right about here. Here we have the sum from n equals one to infinity of the arctangent of x over n squared plus one. How can we figure out if the integral test is something we can use, and if so, what would this look like? Well, what we first need to do is identify what our a sub n is. That's going to be this whole thing right here. Now what I need to do is make my a sub n into a function f of x, and this can be done by taking all of my n's and replacing them with x. So my function f of x is going to be equal to the arctangent of x, and keep in mind the arctangent is just the same thing as the inverse tangent, is going to be over x squared plus one. Now first off, there are three criteria that need to be met for us to be able to use the integral test. The function needs to be positive, continuous, and decreasing on this interval from n equals one to infinity. So in this case, we're looking for values where x is greater than or equal to one, because that's going to infinity. So let's take a look here. First off, is this going to be a positive function? Well, looking at this, you can see that I have my arctangent on top. It's only going to be positive numbers that we put in for this x right here, because notice we go from one to infinity. So that's gonna be all positive numbers, meaning that the top here is gonna be positive. Anything that we put on bottom is gonna be positive as well, and, again, these are all positive numbers we're plugging in anyways. So, yes, this first condition is met. Now next, is this going to be continuous? Well, this is where we get a bit of a question that arises here because notice that we have this rational function. Now since we're looking at these all positive numbers, we have this domain on top here of all real numbers for this arctangent. So because we can go from negative infinity to infinity here, and I can see that we're only focusing on numbers that are greater than or equal to one, that means we are gonna be continuous for this arctangent, but that's not really what I'm focused on. What I'm focused on is this denominator here, the x squared plus one. Is there any case where we're going to break this function? Well, what I know is that the denominator, x squared plus one, cannot be equal to zero because we're not allowed to divide by zero when it comes to rational functions. However, if I go ahead and solve for x here, because we subtract the ones, we get x squared equals negative one, and there's not really much I can do from here. If I try to take a square root on both sides, that's just gonna give me an imaginary number right here. And if you look at this, we're only plugging in values that are greater than or equal to one, so anything we plug in here is just going to give me a positive result on bottom as well. And so because of this, we can recognize that we're not gonna get a case where the denominator equals zero at any point, so we could say that this is indeed continuous. Now next, I need to check if this function is decreasing. Now the way that I can do this is just by thinking about what's going to happen in the long run as we go as we have x values that are greater than or equal to one, and see if this is going to be decreasing for all of those values as we continue to progress towards infinity. Now thinking about this, I noticed that we have this denominator here that's being squared. I can see this x squared in the denominator is going to be growing faster than anything we would plug into this arctangent. And because of that, what that tells me is that this bottom here is blowing up more than the top, and that means that this whole thing is going to be getting smaller and smaller. So, yes, this is decreasing. So because these three criteria are met right here, we can apply the integral test. So what I'm going to do is I'm going to integrate my function f of x. So that's going to be the integral. It's going to be of arctangent of x over x squared plus one integrated with respect to x. Now what are my bounds going to be here? Well, let's take a look at where we're starting. We're starting at one, and we're gonna go up to infinity. But keep in mind, when dealing with integrals, I can't write this infinity as just some boundaries since infinity is not really a number. So what I'm going to do is choose some variable t. So we're going to have the limit as t approaches infinity. And then what I'm going to go ahead and do is write out this integral from one to t of the arctangent of x. It's going to be over x squared plus one integrated with respect to x. And from here, I just need to evaluate this improper integral. So let's do this by the steps here. Well, first off, what I want to do is some kind of variable substitution or something that's going to allow me to get things to cancel or look a little bit more simple. And the way I can do that is by doing a variable substitution on this arctangent of x right here. Because what I can do is say that u is equal to the arctangent of x. Now what exactly would the derivative of this be? Well, you may remember the derivative of an arctangent is actually something that's familiar, because it's the same thing as the derivative of an inverse tangent. That ends up being one over x squared plus one, and we do this with respect to x. So rewriting out this integral will keep this limit as t approaches infinity. That's going to be the integral from one to t. Now on top, we said that this whole arctangent here was u, And in the denominator, we have this x squared plus one with respect to x. Now notice they have a mix of variables here. I have a u, and then I have an x squared. And keep in mind that these bounds up here are still with respect to x. In fact, what I'm actually going to do is I'm going to eliminate these bounds on top here, and we'll just focus on the integral right here. We'll place those bounds in at the end. Now looking at this, notice that we have an x squared plus one d x right here, and this is the same thing as d u. So I can rewrite this whole thing. In this case, and and we'll ignore this, actually, I will keep this limit on because we are gonna apply the balance here. So I'll keep this limit here. But what we're gonna do is we're going to say this whole thing is the integral of u, and then this whole thing just ends up being d u, because the one over x squared plus one would be that x squared plus one in the denominator and then the dx. So now we're just integrating u with respect to u. So doing this, I can just apply the power rule. So keep this limit as t approaches infinity, and then the power rule here, well, the integral of u is just going to be u squared over two using the power rule here. Now what I'm going to go ahead and do from here is I'm going to take my u, which is the arctangent of x right here, I'm going to plug it back in. So we said that u is the arctangent of x, so that means this is the same thing as having the arctangent of x squared over two. And then when I write this, I can go ahead and bound this from my lower bound of one to my upper bound of t. So here we go. Now so now that I've gone ahead and I found this these bounds right here, well, what I now need to do is think about how I can go ahead and apply these. So now that we have our upper bound of t, our lower bound of one, well, let's go ahead and plug this in. So doing so, that's going to give me the arctangent of t and all of this squared over two, plugging in my upper bound of t. I'm also gonna plug in plug in my lower bound of one, which is going to give me, in this case, it's gonna give me the arctangent of one, this whole thing is squared, over two, and then don't forget we need to also have this limit out front, this is an important thing to remember right here. So doing this, well, we're going to have the limit as t approaches infinity. It's gonna be out front. So this is what we get. So now we need to figure out how this is going to behave. And the way that I can do this is by thinking about what's going to happen to my arctangent as we approach infinity. Because notice here that we have this t inside of our arctangent here, and this is going this whole value is being squared. What is this all going to come out to be? So because we have this arctangent of t squared over two, and t is approaching infinity here, well, what this is all going to approach as we go to infinity for our arctangent, as you may recall, this whole thing just comes out to pi over two as we go ahead and follow one of these asymptotes or go to infinity. So because of this, this whole thing we could write as one half because that's that two right there, and this would be pi over two. This is going to be minus, and then we'll have another one half right about here. And the arctangent of one squared, well, that's just going to be the same thing as pi over four squared. And, also, I need to write a square right here since this whole arctangent is squared. Go ahead and finish things off here. What I could do is I could actually simplify this, and all of this is gonna come out to pi squared over eight. You multiply everything out and combine these these denominators and the numerators after squaring this fraction. And then right here, well, that's all going to come out to pi squared over 32. And notice that we would get a finite number that we converge to after doing this improper integral. So because of that, we could say this is convergent. So convergent is going to be the solution to example a, and that's how we could solve this. But now let's go ahead and move to example b. For example b, we are also asked to use the integral test here. So let's go ahead and figure out how we can do this. I'll start by identifying my a sub n, which is right there, and I'm going to turn this into our function f of x. So f of x is going to be one over three x plus two, replacing that n with an x. Now is this positive? Well, looking at this, I can take a look at the interval that we have, and I can see that this is gonna be for x greater than or equal to one. And I can see for all x values that we plug in that are greater than one here or equal to one, this whole thing is going to be a positive fraction at all times. There's nothing that's gonna make this negative, so we can say, yes, this function is positive. Now next, I need to check if this is continuous. Well, to do this, what I need to do is take a look at my denominator right here, three x plus two. And I know that the denominator of a rational function cannot be equal to zero. So go ahead and subtract two on both sides of this equation. The mean that three x cannot be equal to negative two, then I can go ahead and divide both sides of this equation by three. So x here cannot be equal to negative two thirds. But notice here negative two thirds is going to be outside of this value right here. X has to be greater than or equal to one, and notice that this is clearly less than one. So because of that, we don't have anything to worry about, this function is going to be continuous for all x is greater than our starting index. Now next, we'll check if this is decreasing. Well, right off the bat, I actually can see this is going to be a decreasing function, because the only x we have shows up right about there. And because of that, that means the only thing that's going to be growing is our denominator, which means this whole thing is going to be becoming less and less, so you can say, yes, this is a decreasing function. So that means I can now apply my integral test, which is what I'm going to do. So what I'm going to do is integrate, and this is going to be our function f of x. So we're going to have one over three x plus two integrated with respect to x. And my balance will those are gonna go from one to infinity. But But, again, I can't write infinity as this upper bound right here, so I need to go ahead and replace that infinity with a different variable. So you're going to have the limit as t approaches infinity for the integral from one to t of one over three x plus two integrated with respect to x. Okay. Now what do we do from here? Well, from here, it just turns into solving this integral that we see. Now what I'm going to do is a variable substitution. I'm gonna say that my three x plus two here is equal to u, meaning that u is equal to three x plus two. And my derivative here is going to be well, that's just going to end up being three right there because the derivative of two is constant. This is going to be three with respect to x. So writing this all out, we're going to have the limit as t approaches infinity, and we're going to have this integral right here, which is going to be one over u, and that's d x. But notice here that d u is just three d x. What I can do is multiply this and then divide by three and say that that's d x. So what that's ultimately just gonna give me is d u over three. Now what I can do from here is take that three in the denominator and pull it to the front, so that's gonna give me the limit as t approaches infinity of one thirds times the integral of one over u du. And now I just need to go ahead and use a natural log right here, which is going to allow me to find the antiderivative or integral of that one over u. So it's gonna give me the limit as t approaches infinity, and we're gonna have one third. That's gonna be the natural log of u. And then I'm gonna go ahead and take that u and replace it with the u that we had up here, which is my three x plus two. So this is what we get, and then this is all going to be bounded from zero to t, putting those bounds back into place now that we have things back in terms of x. Now from here, I just need to go ahead and evaluate my bounds. So plugging them in will keep this limit as t approaches infinity. I'll start by applying my upper bound of t. Well, that's going to give me in this case, we'll have this one third on the outside here, and it's gonna give me the natural log of three times t plus two. Then we're going to have minus, and then I'll plug in zero. So what that's going to give me, in this case, if I plug in zero, is the natural log, because this is still being multiplied by one third, this whole thing, and that's going to be of three times zero, which is just gonna be zero plus two, so we just have the natural log of two. Now notice when we go ahead and did this improper integral right here, notice what happens. This ends up just being a constant right here, the natural log of two. But this, we have a t in here that's continually growing as it goes to infinity. So because of that, this whole thing is going to blow up to infinity. So we end up having infinity minus just a constant, then all multiplied by one third. But since this is just blowing up to infinity, that means this whole thing, this whole this whole improper integral is just gonna blow up to infinity. So because of that, we could say that this is a case that does not exist. This is not a real number right here, so we would say that this is divergent. So in example b, we got an integral that is divergent, and that is the solution to example b. So this is how you can use the integral test when it comes to solving problems or trying to figure out whether a series is convergent versus divergent. So if the integral is convergent, the series is convergent, and the integral is divergent, then the series is divergent. You just wanna make sure that you're always checking off to see that the integral test will actually apply to your situation, and then you can use this test. So hope you found this video helpful, and let's move on.

Confirm that the integral test applies and then use the integral test to determine convergence of the series. (A) ∑ n = 1 ∞ a r c tan ⁡ n n 2 + 1 \sum_{n=1}^{\infty}\frac{arc\tan n}{n^2+1} (B) ∑ n = 1 ∞ 1 3 n + 2 \sum_{n=1}^{\infty}\frac{1}{3n+2}

(A) The function a r c tan ⁡ ( x ) x 2 + 1 \frac{arc\tan\left(x\right)}{x^2+1} x 2 + 1 a rc t a n ( x ) is not continuous, and so the integral test cannot be used (B) The function 1 3 x + 2 \frac{1}{3x+2} 3 x + 2 1 is positive, continuous, and decreasing for x ≥ 1 x\ge1 x ≥ 1 , and the integral test confirms the series is convergent

(A) The function a r c tan ⁡ ( x ) x 2 + 1 \frac{arc\tan\left(x\right)}{x^2+1} x 2 + 1 a rc t a n ( x ) is positive, continuous, and decreasing for x ≥ 1 x\ge1 x ≥ 1 , and the integral test confirms the series is convergent (B) The function 1 3 x + 2 \frac{1}{3x+2} 3 x + 2 1 is not decreasing, and so the integral test cannot be used.

14. Sequences & Series

Convergence Tests

Calculus

14. Sequences & Series

Convergence Tests

Struggling with Calculus?

Join thousands of students who trust us to help them ace their exams!Watch the first video

Multiple Choice

Confirm that the integral test applies and then use the integral test to determine convergence of the series.
(A) $\sum_{n = 1}^{\infty} \frac{a r c \tan n}{n^{2} + 1}$ (B) $\sum_{n = 1}^{\infty} \frac{1}{3 n + 2}$

(A) The function $\frac{arc\tan\left(x\right)}{x^2+1}$ is positive, continuous, and decreasing for $x\ge1$ , and the integral test confirms the series is divergent

(B) The function $\frac{1}{3x+2}$ is positive, continuous, and decreasing for $x\ge1$ , and the integral test confirms the series is convergent

(A) The function $\frac{a r c \tan (x)}{x^{2} + 1}$ is positive, continuous, and decreasing for $x is greater than or equal to 1$ , and the integral test confirms the series is convergent

(B) The function $\frac{1}{3 x + 2}$ is positive, continuous, and decreasing for $x is greater than or equal to 1$ , and the integral test confirms the series is divergent

(A) The function $\frac{arc\tan\left(x\right)}{x^2+1}$ is not continuous, and so the integral test cannot be used

(B) The function $\frac{1}{3x+2}$ is positive, continuous, and decreasing for $x\ge1$ , and the integral test confirms the series is convergent

(A) The function $\frac{arc\tan\left(x\right)}{x^2+1}$ is positive, continuous, and decreasing for $x\ge1$ , and the integral test confirms the series is convergent

(B) The function $\frac{1}{3x+2}$ is not decreasing, and so the integral test cannot be used.

Verified step by step guidance

Step 1: Understand the Integral Test. The Integral Test states that if a function f(x) is positive, continuous, and decreasing for x ≥ 1, and f(n) corresponds to the terms of a series ∑a_n, then the convergence or divergence of the series can be determined by evaluating the improper integral ∫f(x)dx from 1 to ∞.

Step 2: For part (A), analyze the function f(x) = arctan(x)/(x^2 + 1). Verify that f(x) is positive, continuous, and decreasing for x ≥ 1. Positivity is clear since arctan(x) and x^2 + 1 are positive for x ≥ 1. Continuity is satisfied because both arctan(x) and x^2 + 1 are continuous functions. To check if f(x) is decreasing, compute the derivative f'(x) and confirm that it is negative for x ≥ 1.

Step 3: Apply the Integral Test to part (A). Evaluate the improper integral ∫[arctan(x)/(x^2 + 1)]dx from 1 to ∞. If the integral converges, the series ∑arctan(n)/(n^2 + 1) converges; otherwise, it diverges.

Step 4: For part (B), analyze the function f(x) = 1/(3x + 2). Verify that f(x) is positive, continuous, and decreasing for x ≥ 1. Positivity is clear since 3x + 2 is positive for x ≥ 1. Continuity is satisfied because 3x + 2 is a continuous function. To check if f(x) is decreasing, compute the derivative f'(x) and confirm that it is negative for x ≥ 1.

Step 5: Apply the Integral Test to part (B). Evaluate the improper integral ∫[1/(3x + 2)]dx from 1 to ∞. If the integral converges, the series ∑1/(3n + 2) converges; otherwise, it diverges.