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### Course: Calculus, all content (2017 edition)>Unit 7

Lesson 8: Basic convergence tests

# Integral test

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.

## Want to join the conversation?

• 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?
• 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.
• Wait, how do you know when to use the integral test?
• 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.
• 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
• Sure, you can try to estimate the integral however you like. Remember when we were doing Reimann sums? There's no god of math or anything mandating that you do a left Reimann sum and not a right Reimann sum. In this case, we could use a pretend left Reimann sum, and figure out that our infinite series sums to something greater than the integral of [ 1/n² ] from 0 to ∞, but that wouldn't help. What we want is, will the sum be a finite number? And in order to answer that, we have to say, will the sum be less than some finite number? We found that the sum is less than two, so we know that it converges.
(1 vote)
• 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?
• 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.
• how did u integrate from 1 to infinty when you are summing from 2 to infinity
• 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 :) .
• why is the sum of 1/n unbounded? doesn't it converge as well?
• 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.
• 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.
• There is an identity that the sum shown above converges to pi^2/6. What is the proof of it?
• 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?
• Sal sometimes mentions some proofs are not rigorous, I'm wondering whether it is because the rigorous proofs are harder to understand at this level or they take more time to prove. And are these not-so-rigorous proofs taught in standard AP calculus classes too? Cheers.
• Rigorous proofs aren't necessarily harder (as something being hard is subjective). However, you do need more knowledge to rigorously prove something, and a rigorous proof is one that leaves no mathematical ambiguity. I'll give you a simple example. You must have learnt that the surface area of a sphere is 4πr^(2). Now, when you're first introduced to this topic, a teacher usually explains this formula by saying that a sphere is four circles, and as the area of each circle is πr^(2), the area of four would be 4πr^(2).

Now, this proof isn't rigorous. Imagine cutting a paper sphere into four circles by hand. You'd have a lot of waste paper (which would also contribute to the area) and your circles needn't be congruent. So, this "proof" isn't very acceptable.

Now, for a more rigorous proof, we turn to calculus (Won't go into too much detail as it requires knowledge of multivariate calc). If you're able to parametrize a sphere (which isn't too hard), a surface integral will give you the surface area. This is actually where that formula comes from, but as you obviously can't teach calculus to sixth graders, they go with a more loose proof and emphasize the formula more.

So, see how the non rigorous proof involved basic geometry but the more rigorous one involved multivariate calc. There might be an even more rigorous one if you dive deeper, but yeah, that's where my knowledge is bound for now. All in all, rigorous proofs do need more knowledge of the subject, but they aren't necessarily hard. They're just a bit more aggressive with the nitty-gritties

As for your last question, I'm pretty sure I read that AP classes do not focus on proofs at all. However, KA does them because you can always learn something new from proofs. Personally, I love proofs a lot. THey hold the essence of the formula.