### Video Transcript

Using the integral test, determine
whether the series one over two multiplied by the natural log of two add one over
three multiplied by the natural log of three add one over four multiplied by the
natural log of four and so on is convergent or divergent.

Letβs begin by reminding ourselves
of what the integral test is. Suppose that π of π₯ is
continuous, positive, and decreasing for all values of π₯ greater than or equal to
some value π and π of π equals π π. Then if the integral between π and
β of π of π₯ with respect to π₯ is convergent, then so is the series the sum from
π equals π to β of π π. If the integral between π and β of
π of π₯ with respect to π₯ is divergent, then so is the series the sum from π
equals π to β of π π.

Note that both of these integrals
are improper integrals. Thatβs because one of the limits of
integration is β. We deal with improper integrals by
replacing the infinite limit with a variable and then evaluating the integral as
that variable approaches β. If we then find that when we
evaluate the integral we get β, then the integral is divergent. Otherwise, if we get a finite
number, then the integral is convergent.

Letβs start by finding a way to
express this series in Ξ£ notation. We can see that it has the form the
sum of one over π multiplied by the natural log of π. But weβve got to spot that this
runs from π equals two to β. So our π π is one over π
multiplied by the natural log of π. So the function π of π₯ that weβre
going to be integrating is one over π₯ multiplied by the natural log of π₯. And our limits of integration are
two and β. In order to use the integral test,
we must check that weβve met the conditions that π of π₯ is continuous, positive,
and decreasing on the interval.

So letβs start with checking if
this function π of π₯ is continuous. We need to make sure that this is
continuous from π₯ is greater than or equal to two. And because the quotient of two
continuous functions is continuous and π₯ multiplied by the natural log of π₯ is
defined for all π₯ greater than or equal to two and is not equal to zero on our
interval, then one over π₯ multiplied by the natural log of π₯ is continuous. We also have that all values of one
over π₯ multiplied by the natural log of π₯ are positive for π₯ greater than or
equal to two. And finally, as the natural log of
π₯ is an increasing function, so is π₯ multiplied by the natural log of π₯. So one over π₯ multiplied by the
natural log of π₯ is a decreasing function. So weβve satisfied all of the
conditions required for the integral test.

So remember weβre trying to
determine the convergence of this integral. And as this is an improper
integral, weβre going to rewrite this as a limit. This is the limit as π‘ approaches
β of the integral between two and π‘ of one over π₯ multiplied by the natural log of
π₯ with respect to π₯.

Letβs evaluate this integral using
integration by substitution. If we start with the indefinite
integral one over π₯ multiplied by the natural log of π₯ with respect to π₯, then
using the substitution π’ equals the natural log of π₯ so that dπ’ by dπ₯ equals one
over π₯. And remember that dπ’ by dπ₯ is not
a fraction. But we do manipulate it a little
bit like a fraction when weβre doing integration by substitution. So this rearranges to give us that
dπ’ equals one over π₯ dπ₯. Then we can replace one over the
natural log of π₯ with one over π’. And we can replace the one over π₯
dπ₯ with dπ’.

At this point, we recall the
general result that tells us that the integral of one over π’ with respect to π’ is
the natural log of the absolute value of π’. And because here weβre doing
indefinite integration, we add a constant of integration π. But because weβre actually looking
for the integral between two and π‘ of one over π₯ multiplied by the natural log of
π₯ with respect to π₯, we need to substitute back in π’ equals the natural log of
π₯. This gives us the limit as π‘
approaches β of the natural log of the absolute value of the natural log of π₯.

And remember we need to evaluate
this between two and π‘. This gives us the limit as π‘
approaches β of the natural log of the absolute value of the natural log of π‘ minus
the natural log of the absolute value of the natural log of two. But the natural log of the absolute
value of the natural log of π‘ approaches β. So evaluating this whole expression
with the limit as π‘ approaches β just gives us β. This result means that the improper
integral between two and β of one over π₯ multiplied by the natural log of π₯ with
respect to π₯ is divergent. And the integral test tells us that
if we find this integral to be divergent, then this series is also divergent. So by the integral test, this
series is divergent.