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## Basic convergence tests

# Worked example: Integral test

AP.CALC:

LIM‑7 (EU)

, LIM‑7.A (LO)

, LIM‑7.A.6 (EK)

## 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.