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### Course: AP®︎/College Calculus BC > Unit 10

Lesson 4: Integral test for convergence# Worked example: Integral test

See how the integral test is put to use in determining whether a sequence converges or diverges.

## Want to join the conversation?

- I'm confused about part 2 of the integral test. In the last video, Sal showed that the series is an underestimate of the integral, and therefore a convergent (finite) integral would bound the series and make the series convergent too.

But it doesn't follow that just because the integral is divergent, then the series also diverges, because the integral is larger than the series.(85 votes)- Actually, for part 2 you set it up so that the integral is
**smaller**than the series. Therefore if the integral diverges, the corresponding series must diverge as well. It's all in how you draw your rectangles for the series (left endpoints or right endpoints) which allows you to visualize the series as being either less than or greater than a desired (and known) integral. Does that help?(80 votes)

- Usually the videos are very clear about everything, but this is the first time i was left a little unsure.

In the previous video ("Integral test intuition") we see Sal using the (1/x^2) function as an upper limit to the sum of (1/n^2) and showing how each block in the sum is less than its respective section in the integral. I am assuming that when n=1, the point on the graph is from ( n, (1/n^2) ) and drawn left to previous point. The boxes are clearly smaller and thus this "proof" is clear.

For this video however, without any shown reasoning - other than the function is now (1/x) and the sum is the harmonic series, 1/n, Sal draws from (n, 1/n) but to the right making the boxes bigger. Thus the sum is bigger and is pushed to divergence etc. If I hadn't accepted the rule that diverging(converging) function = diverging(converging) series respectively, then I would draw the boxes to the left (as in the previous film) making them smaller than the function and thus the whole test inconclusive.

I'm sure there is a small detail somewhere about where the integral or sum starts etc, but since i didn't notice it here, it will be a trap during an exam.

Looking forward to your answer.(52 votes)- I think that Sal isn't worried about that detail because the functions are so closely connected that it doesn't matter whether you overestimate or underestimate the function. Both the function and the estimation converge or diverge independent of the estimation. The only reason estimating over or under would matter would be to find the value that it approaches, but because they are both estimations that follow the function closely and do not present a number that's accurate, the integral itself is all that matters.

Sal does show some proof in the first video by comparing that sum to the integral plus the first value of the series. ∑ < ∑(1) + ∫ This allows comparison to an overestimate and allows a function that converges to be proven as convergent.

In the second video, Sal compares the sum directly to the integral ∑ > ∫ leaving the integral in a position where it is smaller than the sum and and where it is able to prove the sum divergent.

This could be done every time you use the integral test but the integral follows the series so closely that it would be a waist of time. Correct me if I'm wrong.(10 votes)

- In the sketches, how can you use left Riemann sum for 1/x and then right Reimann sum for 1/x^2? Isn't the Riemann sum open to interpretation, and not actually prove anything in this situation? My question is: why don't you be consistent with either the right or left hand Riemann sum?(21 votes)
- In the previous video titled "integral test intuition", Sal uses a right Riemann sum for 1/x^2 because it represents the sum of rectangles that lie entirely underneath the curve. The purpose of doing that was to argue that the sum of the first n terms minus the first term is less than the integral from 1 to n. And since the improper integral was convergent in that case, the series was bounded above by a positive number. And since we have an increasing series, that series must converge to some number.

In this video, Sal uses left Reimann sums because the rectangles lie above the curve. This again allows him to convincingly argue that the sum of the series(1/n) is divergent because the for any given n, the sum of the first n-1 terms is always greater than the integral between 1 and n. And since the integral of f is divergent, and f is always positive, the integral must be unbounded. That implies that the series is unbounded and must therefore diverge.

In short, Sal's decision to switch from a right to a left Riemann sum was strategic on his part because the left Reimann sums for 1/x always lie above the curve, making it easier for him to visually confirm that it does indeed diverge.(7 votes)

- What's the difference between this function and the one in the previous video (other than 1/x^2 and 1/x) where he drew the area of 1 from 0 to 1, but here he drew the area of 1 from 1 to 2?(9 votes)
- there is something strange with limits of int(1/x) and Int(1/x^2. looking at graphs of antiderivatives its obviuos that the first one diverges and the second one converges on (1, infinity). But if I observe graphs of d(lnx)/dx and d(-1/x)/dx and areas under those curves ( integrals we initially have ) , then it looks like both areas should converge. Farther from 1 is x, smaller f(x) and therefore area is increasing/decreasing with smaller and smaller portions, which eventually should lead to convergence. why does it seem to be true ? What am I getting wrong?(6 votes)
- If understand your question, your concern is that even though 1/x, and 1/x^2 have a similar looking shape when graphed, why then do their series not converge?

You cannot always tell weather the series of a function converges based on its graphed shape alone, which is why we need all the tests in this playlist. You must analyze the function, it's an unpleasant truth.(4 votes)

- Simply can anyone say me how series and integral are related?(3 votes)
- Although integrals are often associated with the antiderivative, which to be fair is how we calculate them, the main definition and central application of a definite integral is area under a curve during a certain interval. Like infinite series, improper integrals--or integrals which have unbounded intervals--also converge or diverge. Using right hand Riemann sums as a way to represent out infinite series, the improper integral will always be an overestimate of the infinite series. Therefore, if the improper integral converges, the infinite series will also converge. If some of the terms seem vague, check out these links:

Improper integrals: https://www.khanacademy.org/math/integral-calculus/indefinite-definite-integrals/improper-integrals-tutorial/v/introduction-to-improper-integrals

Riemann Sums: https://www.khanacademy.org/math/integral-calculus/indefinite-definite-integrals/riemann-sums/v/simple-riemann-approximation-using-rectangles(5 votes)

- Do we have a simple and actual proof for this test?(3 votes)
- Sal sketches what the proof would look like in two parts, with part 1 (convergence of the integral implies convergence of the series) being in the video preceding this one and part 2 (divergence of the integral implies divergence of the series) being toward the end of this video. While the presentation isn't rigorous enough to qualify as a full proof, we should be able to understand from this explanation why this test has to be true.(6 votes)

- Why did khan start the series from 1 instead of 0 as he did in the previous video (Integral test intuition.) If the first term of the series starts from 0, then it seems like the series will be bounded by the integral of 1/x from 1 to infinity.(3 votes)
- I think it is because 1/x hits a vertical asymptote at x=0, which means it is going towards infinity in the vertical direction. and that makes testing this series considerably harder.(4 votes)

- why does this only work if the f(x) is decreasing and positive? wouldn't it also work if f(x) is increasing and negative?(4 votes)
- it would, but if you just pull the minus sign out of the function you get a decreasing positive function, so it's essentially the same condition(1 vote)

- Surely the harmonic series, which was proven earlier to be divergent, fits the criteria to be convergent under this test?(4 votes)
- The harmonic series was also proven to be divergent in this video by the integral test because the integral of 1/x is ln(x) which goes to infinity. Sorry for the multiple posts, I think I deleted all the extra copies, my arm was resting on the mouse haha.(0 votes)

## Video transcript

- [Voiceover] Let's now
explain to ourselves, I guess you could say, a more formal communication of the Integral Test. Integral Test. So it tells us that if we
assume that we have sum f of x, if we have sum f of x that is positive... positive, continuous... continuous... continuous, and decreasing... and decreasing... and decreasing on some interval, on... So starting at k, and including
k, all the way to infinity. Then we can make one of two statements. We could say either that
if the improper integral from k to infinity of f
of x dx is convergent, is convergent, then... then the sum, the infinite
series from n is equal to k to infinity of f of n, is also convergent. Is also convergent... Convergent. And this is actually the
case that we saw when we looked at one over n squared, but I'll look at that in a second. But the second claim that we could make, or the second deduction that we might be able to make using the Integral Test, is if that's the other way around. That if the integral from k to infinity, the improper integral of
f of x dx, is divergent, divergent, then the same thing is true for the corresponding infinite series. Then this infinite series right over here is also going to be... is also divergent. And as I already mentioned,
in the last video we already saw this in the case of f of x is equal to one over x squared. We saw that since the
integral from one to infinity of one over x squared,
over one over x squared dx, is convergent, in fact, it equals one. It equals one. Because of that, we were
able to say that the sum from n equals the sum
from n is equal to one to infinity of one over n
squared, is also convergent. Also convergent. And now we can see an example
where we go the other way. For example, we know that this integral... Let me write the integral down. Let's start with this integral. From one to infinity,
not of f of x is equal to one over x squared, but let's say that f of x is equal to one over x. Actually let me just write that down. Let's just start with f of
x is equal to one over x. It definitely meets our
conditions that it is positive, and let's say we were going to consider it over the interval... over the interval from one to infinity. So it meets this first constraint. Over this interval,
one over x is positive, it is continuous, and it is decreasing. As x increases, f of x decreases. So the Integral Test should apply. So let's see what the improper integral from one to infinity of this would be. So if we take... If we go from one to
infinity of one over x, of one over x dx, this is equal to... We could write this as the limit as t approaches infinity
of the definite integral from one to t, of one over x dx, which is equal to the limit,
as t approaches infinity, of, take the antiderivative, is going to be of the natural log of x, the natural log of x going from one to t. One to t, we could do the... Well, it's really the absolute value of x, but we're dealing with positive x's here so it's just going to
be the natural log of x, which is going to be the
limit as t approaches infinity of the natural log of t, or I could even say the natural log of the absolute value of t, which is going to be the natural log of t,
because it's positive t's, minus the natural log of one. Minus the natural log of
the absolute value of one. The natural log of one
is zero, so it's going to be the natural log of t. The limit of that approaches infinity. But the limit as that approaches infinity is just going to be unbounded, this is going to go to infinity. This right over here is divergent. So this right over here is divergent. So this is divergent. And because this is
divergent, we can then say, we can then say, by the Integral Test, we can then say the Integral Test... Once again, our function
over this interval, positive, continuous, decreasing. We saw that this improper
integral right over here is divergent, and then by the second point of the Integral Test we can say therefore, and I haven't rigorously
proved it yet, but hopefully I gave you a good intuitive justification in the previous video,
that the integral... that the infinite series
from n equals one to infinity of one over n, which
is the harmonic series, that this is also, this is also... this is also divergent. So we've already shown
that the harmonic series is divergent using that
very beautiful, elegant proof by Oresme, I think
I'm probably mispronouncing his name, though used the
comparison test, but just like this we have used the Integral Test to show that it is also divergent. And once again, let's
remember what the whole motivation of the Integral Test is. Let me draw f of x is equal to one over x. So f of x is equal to one
over x would look like... Do my best attempt here. So let's say that's one, two, three, and that is one, two... And so let's see, when
x is one f of x is one, when x is two f of x is 1/2 or 1/3. If it's 1/2 here, it
will be two over here. So it looks like this. So this is f of x is equal to one over n, and once again we see that
then over the interval we care about, from one to
infinity, it's definitely positive, continuous, and decreasing. And if we look at this
sum right over here, we could view this sum as... Let's do that, let me write it down. So the sum... The sum from n equals one
to infinity of one over n, is equal to one plus 1/2
plus 1/3, and of course we keep going on and on and on and on. In this case, since we want
to show it's divergent, we could say, "Hey, look,
this is an overestimate of, "of this area here." Let me be clear. So we have this area. We have this area in green, which is what the improper
integral is denoting. So that right over there
is the improper integral from one to infinity of one over x dx. Now you could view this as an overestimate of that area. So this first... This one right over here, you could say that this is this one
height times one width, so that's that block right over there, that's the area of that... is going to be equal to one. Then this over here,
1/2, you could view that as the area of the next block, of the next block. So you could kind of view this
as a left-sided Riemann sum, I guess is one way to think about it. And so this is going to be 1/2... Yeah, left-sided Riemann sum. So this is going to be 1/2, and then the 1/3 is going to be this one, it's going to be this one. I noticed they're all... The actual area we care
about, the improper integral, it's all contained in these blocks. So this is going to be an over... is going to be larger than
this, but we've already seen that this is
unbounded towards infinity. This is divergent. So if this is larger than this, and this is divergent,
this goes to infinity, then this must also go to infinity. So that's exactly where the
Integral Test is coming from.