AP®︎/College Calculus BC
The integral test helps us determine a series convergence by comparing it to an improper integral, which is something we already know how to find. Learn how it works in this video.
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- So if you're continuously adding something, how is the sum decreasing? I know you can take the derivative and see if it's positive or negative, but is there any other justification that would make it click?(4 votes)
- I hope this is an answer to your question (as I understood it to be):
The sum isn't decreasing, but each successive term in the series is. So what can we say about that? Let’s take a look at the behavior of 1/n². The first term is 1/1²=1, the next is 1/2²=1/4, the next is 1/3²=1/9, then 1/4²=1/16 . . . . . 1/100² = 1/10000 etc. So you can see that, yes, the terms are positive, and the are getting smaller quite fast. But still you wonder, how can adding positive numbers to a sum ever stop the sum from growing bigger and bigger and bigger. Here is the general idea, and while the example I am about to use is not the sequence above, the same logic applies. Have you heard of Zeno's paradox?
Zeno argued that motion was inherently impossible. Suppose you wish to travel a distance, d. Well to do so you would first need to travel half the distance to d first, but to travel half the distance to d you would first have to travel "half of the half" of the distance to d, and then "half of the half of the half" This argument goes on forever, so it appears that the distance d cannot be traveled.
The situation is somewhat similar here. Each successive term of the series goes a "sufficient distance less"* than that of its predecessor that it is just like the example of moving half the distance more each time. So even though you are continuously adding to the sum, the values are getting so small that you will never fully cover the distance, d. You can get as close as you want to d, but each time you are almost there, you can only travel half the remaining distance, thus the series limit, d, is forever beyond your reach.
* This behavior only happens if each successive term is small enough. For example, 1/n diverges, 1/n² does not.(54 votes)
- Wait, how do you know when to use the integral test?(15 votes)
- When asked to show if a series is convergent or divergent you might spot that such series is "mimicked" by a positive, decreasing and continuous function (there's no fixed rule, you have to train your mind to recognize these patterns). If that is the case you can use the integral test to say something about the series and back it up properly.(3 votes)
- I don´t understand why in the area of f(x)=1/x^2 Mr.Khan draws the area of the first term (1) to the left an in the area of f(x)=1/x he draws it to the right.
Anyway, it was really helpfull. Thanks(21 votes)
- So if the area under the curve is finite then the series will converge, if the area under the curve is infinite then the series will diverge. Or am I missing something here?(7 votes)
- That's correct; it's also worth noting that the area under the curve will not necessarily be equal to the sum of the series, but the infinite series and the improper integral will either converge together or diverge together.(7 votes)
- how did u integrate from 1 to infinty when you are summing from 2 to infinity(6 votes)
- When you take the sum of the series from 2 to infinity, you include the green area under the curve
1/(x^2)(which has an area of 1/4) as that is the first term in the sum. To include the same green area in the integration, we have to integrate from 1 to infinity.
Hope this helped :) .(4 votes)
- 1:09why is the sum of 1/n unbounded? doesn't it converge as well?(5 votes)
- The series 1/n does not converge, even though it slowly decreases it is not enough to make it converge. The series 1/n^2 in the other hand does converge beacuse it decreases way faster. Sal has a video explaining a proof for the divergence of this series, the Harmonic Series. Here it is: https://www.khanacademy.org/math/calculus-home/ap-calculus-bc/series-convergence-bc/convergence-divergence-tests-bc/v/harmonic-series-divergent(5 votes)
- There is an identity that the sum shown above converges to pi^2/6. What is the proof of it?(4 votes)
- Not too hard to find - try this first and follow where it leads you.
- Why can't the conditions for the integral test be negative, increasing, and continuous, instead of positive, decreasing, and continuous? What is the significance of each of the conditions?(3 votes)
- If given a diverse set of series to test for divergence/convergence, how would we know when to use the Integral Test? Are there characteristics you can look for in a series that indicate the Integral Test may be the way to go?(3 votes)
- Why didn't Sal just do the improper integral from 0 to infinity instead of going from 1 to infinity and just adding 1. Was it just a small aesthetically pleasing thing, or was there some concrete mathematical reasoning behind it.(2 votes)
- Because the integral from 0 to ∞ of x^-2 doesn't converge. So if he really wanted an upper bound on the series, he had to make the lower bound of the integral some positive number. Given the way the series started, 1 was most convenient.(2 votes)
- [Voiceover] Let's explore a bit the infinite series from n equals one to infinity of one over n squared. Which of course is equal to one plus one fourth, that's one over two squared, plus one over three squared, which is one ninth, plus one sixteenth and it goes on and on and on forever. So there's a couple of things that we know about it. The first thing is that all of the terms here are positive. So all of the terms here are positive. So they're all positive and that they're decreasing. It looks like they're decreasing quite quickly here from one to one fourth to one ninth to one sixteenth, and so they're quickly approaching zero, which makes us feel pretty good that this thing has a chance of converging. And because they're all positive we know that this sum right over here, if it does converge is going to be greater than zero. So the only reason why it wouldn't converge is if somehow it goes unbounded towards infinity, which we know if this was one over n it would be unbounded towards infinity. So this says that's a possibility here. So if we could show that this is bounded, then that will be a pretty good argument for why this thing right over here converges because the only reason why you could diverge is if you went to either positive infinity or negative infinity. We already know that this thing isn't going to go to negative infinity because it's all positive terms. Or you could diverge if this thing oscillates, but it's not going to oscillate because all of these terms are just adding to the sum, none of them are taking away because none of these terms are negative. So let's see if we can make a good argument for why this sum right over here is bounded, especially if we can come up with the bound, then that's a pretty good argument that this infinite series should converge. And the way that we're going to do that is we're going to explore a related function. So what I wanna do is I wanna explore f of x is equal to one over x squared. You could really view this right over here one over n squared as f of n if I were to write it this way. So why is this interesting? Well let's graph it. So that's the graph of y is equal to f of x. And notice this is a continuous, positive, decreasing function, especially over the interval that I care about right over here. I guess we could say for positive values of x, it is a continuous, positive, decreasing function. And what's interesting is we can use this as really an underestimate for this area right over here. What do I mean by that? Well one, this first term right over here, you could view that as the area of this block right over here. That is f of n or I guess you can say f of one high and one wide, so it's going to be one times one over one squared or one. Let me make sure I'm using different colors. This term right over here, that could represent the area of this block, which is one fourth high and one wide so it is going to have an area of one fourth. What could this one represent? Well the area of the next block if we're trying to estimate the area under the curve. And this might look familiar from when we first got exposed to the integral or even before we got exposed to the integral and we were taking Riemann sums. So that right over here, that area, is going to be equal to one ninth. So what's intriguing about this is we know how to find the exact area, or the exact area from one to infinity, from x equals one to infinity. So maybe we can use that somehow. We know what this area is right over here, which we can denote as the improper integral from one to infinity of f of x dx. We know what that is and I'll figure it out in a little bit. And if we know what this is, if we can figure out the value that's going to be an upper-bound for one fourth plus one ninth plus one sixteenth on and on and on and on. And so that will allow us to essentially bound what this series evaluates to and as we said earlier that would be a very good argument for its convergence. So the whole point here, I'm not doing a rigorous proof, but really getting you the underlying conceptual understanding for a very popular test for convergence or divergence which is called the integral test. Let me just write that down just so you know what this is kind of the mental foundations for. So what do I mean here? So let me write this sum again. Let me write it a little bit different. So our original series from n equals one to infinity of one over n squared. It's going to be equal to this first block, the area of this first block plus the area of all the rest of the blocks, the one fourth plus one ninth plus one sixteenth, let me do this in a new color. Which we could write as the sum from n equals two to infinity of one over n squared. So I just kind of expressed this as a sum of this plus all of that stuff. Now what's interesting is that this, what I just wrote in this blue notation that's this block plus this block plus the next block, which is going to be less than this definite integral right over here. This definite integral, notice it's an underestimate it's always below the curve, so it's going to be less than that definite integral. So we can write that this thing is going to be less than one plus instead of writing this I'm gonna write the definite integral. One plus the definite integral from one to infinity of one over x squared dx. Now why is that useful? Well we know how to evaluate this and I encourage you to review the section on Khan Academy on improper integrals if this looks unfamiliar, but I'll evaluate this down here. We know that this is the same thing as the limit as, I'm going to introduce a variable here, t approaches infinity of the definite integral from one to t of and I'll just write this as x to the negative two dx. Which is equal to the limit as t approaches infinity of negative x to the negative one, or actually I could write that as negative one over x. And we're going to evaluate that at t and at one, which is equal to the limit as t approaches infinity of negative one over t and then minus negative one over one so that would just be plus one. And as t approaches infinity this term right over here is going to be zero, so this is just going to simplify to one. So this whole thing evaluates to one. So just like that we were able to place an upper-bound on this series. We're able to say that the series under question or in question, so the infinite sum from n equals one to infinity of one over n squared is going to be less than one plus one or it's going to be less than two. Or another way to think about it, it's going to be the two is this area, that's one right over there plus this area right over here. So we're saying that this sum is going to be less than two so we have bounded it above. So we know that it cannot go to positive infinity. Because all the terms are positive it's definitely not going to go to negative infinity. And because all the terms are positive we also know that this isn't going to oscillate between two different values, so this gives us a pretty good sense that this series converges. And the logic we just used here to argue for why this converges, once again not a rigorous proof, but this is the underlying logic of the integral test.