### Video Transcript

In this video, weβll learn how to
find the condition for which the π-series converges. And weβll prove the divergence of
the harmonic series using the integral test. Weβll see some examples of the
application of the condition for the π-series to determine whether a series
converges or diverges.

Letβs firstly remind ourselves of
the difference between a convergent series and a divergent series. If weβre trying to determine
whether a series is convergent, weβre trying to find out whether the sum of the
terms gets closer and closer to a specific value as you increase the number of
terms. Otherwise, we have a divergent
series, which means the series does not have a finite limit. So this could be a series that
tends to positive or negative β when you add an infinite number of terms
together.

Thereβs a really important series
called the harmonic series, which is an infinite series of the form the sum from π
equals one to β of one over π which gives us one add one over two add one of
the three add one over four, and so on. And this series actually got its
name from its connection to music. A vibrating string produces
wavelengths in proportion to one, one over two, one over three, one over four, and
so on of the strings fundamental wavelength. But what weβre going to look at is
whether the harmonic series is convergent or divergent. And weβre going to prove this using
the integral test.

Recall, the integral test says that
if we define a function π from a series, which is a continuous positive and
decreasing function on the interval between one and β, then if the improper
integral of the function is convergent, then the series is convergent. If the improper integral is
divergent, then the series is divergent. When the limits of this improper
integral exist, the integral is convergent. And when the limit of the improper
integral does not exist or is infinite, the integral is divergent. So letβs now apply the integral
test to our series.

If we start by defining π of π₯ to
be one over π₯, then the integral weβre going to evaluate is the integral between
one and β of one over π₯ with respect to π₯. This is an improper integral
because we have an infinite limit. And we deal with improper integrals
by replacing the infinite limit with a variable and then taking the limits of the
integral as that variable approaches β. And so we can rewrite this integral
as the limit as π‘ approaches β of the integral between one and π‘ of one
over π₯ with respect to π₯. We know a 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₯ at a constant of integration.

And of course, when evaluating a
definite integral from one to π‘, the constant will be subtracted from itself. So we get the limit as π‘
approaches β of the natural log of the absolute value of π₯. This gives us the limit as π‘
approaches β of the natural log of the absolute value of π‘ minus the natural
log of the absolute value of one. But the natural log of one is
zero. So we just need to evaluate the
limit as π‘ approaches β of the natural log of the absolute value of π‘. And we know this limit to be
β. And so this improper integral is
divergent. So by the integral test this series
is divergent. So even though weβre going to be
adding a smaller and smaller number each time, the partial sums are still growing
without bound.

The harmonic series is actually a
specific case of a more general series called a π-series. A π-series is a series of the form
the sum from π equals one to β of one over π to the π power, such as a
series the sum from π equals one to β of one over π squared. And actually, the harmonic series
is just a π-series with π equal to one. Weβre going to have a look at how
we can tell if a π-series converges or diverges. We can do this using the test for
divergence and the integral test. Weβre going to break it down into
different values for different cases of π. Iβll start by writing up the test
for divergence.

Remember that we know when π is
equal to one, we have the harmonic series, which is divergent. And weβll, firstly, consider the
case when π is equal to zero. Then the limit as π approaches
β of one over π to the π power is just going to be one. As this is not equal to zero, by
the test for divergence when π is equal to zero, the series is divergent. What about the case when π is
strictly less than zero? Well, taking the limit as π
approaches β of one over π to the π power is the same as taking the limit
as π approaches β of π to the negative π power.

Remember here, that if π is less
than zero negative π is greater than zero. And because π represents the
counting numbers for the series terms, π is going to be an integer and so π to the
negative π power will get larger and larger. So as π approaches β, this
limit approaches β. And so, by the test for divergence,
when π is less than zero, this series is divergent. Letβs now look at the case when π
is strictly greater than zero. To do this, weβll apply the
integral test. And here weβre assuming that π is
not equal to one because that would make it a harmonic series. And we already know that that
diverges. So if we define a function one over
π₯ to the π power, then we evaluate the improper integral from one to β of
one over π₯ to the π power with respect to π₯.

And letβs, firstly, rewrite this as
a negative exponents π₯ to the negative π power. And we take our usual approach to
evaluating improper integrals by finding the limit as π‘ approaches β of the
integral between one and π‘ of π₯ to the negative π power with respect to π₯. And using the general rule that the
integral of π₯ to the power π with respect to π₯ is π₯ to the power of π add one
over π add one plus a constant of integration. So we find our integral to be the
limit as π‘ approaches β of π₯ to the negative π add one power over negative
π add one. And remember, this is okay because
weβre assuming that π is not equal to one. And then we substitute in our
limits for integration and then take out one over one minus π as a factor.

From here, we know that one raised
to the power of negative π add one is going to give us one. And also we can write π‘ to the
power of negative π add one and its fractional form as one over π‘ to the power of
π minus one. And now thereβs two cases we need
to consider here, if π is greater than one and if π is less than one. Well, if π is greater than one,
then π minus one is going to be positive. So as π‘ approaches β, π‘ to
the π minus one power also approaches β. So one over π‘ to the π minus one
power approaches zero. So if this term here approaches
zero, then what we have left is one over one minus π multiplied by negative one,
which is negative one over one minus π which is the same as one over π minus
one. And because this limit exists, then
the improper integral is convergent. And so, by the integral test, the
series is convergent for π greater than one.

And now, we look at the case when
π is less than one. If π is less than one, then π
minus one is less than zero, so one over π‘ to the π minus one power, which is the
same as π‘ to the negative π minus one power in a negative exponents form. Note that because π minus one is
negative, then negative π minus one is positive. And so π‘ to the negative π minus
one must approach β as π‘ approaches β. Because this is an infinite limit,
the improper integral, the integral from one to β of π₯ to the negative π
power with respect to π₯, must diverge when π is less than one. And so by the integral test, this
series is divergent for π less than one.

So we found this series converges
only when π is greater than one. So Iβm going to clear some space to
write our conclusion. So we conclude that the π-series
the sum from π equals one to β of one over π to the π power, is convergent
if π is greater than one and divergent if π is less than or equal to one. Note that this is less than or
equal because, remember, we already found the harmonic series when π is equal to
one to be divergent. Letβs now have a look at some
examples of how we can apply this condition to some π-series.

Determine whether the series the
sum from π equals one to β of one over π to the fourth power converges or
diverges.

Well, we recognise this sum to be
the form of a π-series. So we can start by writing out the
condition for convergence of a π-series. The π-series the sum from π
equals one to β of one over π to the π power is convergent if π is greater
than one and divergent if π is less than or equal to one. So for this series, π is equal to
four, which is greater than one. So we can conclude that this series
converges.

Determine whether the series the
sum from π equals one to β of one over π to the power of two-fifths
converges or diverges.

We start by recognising that this
sum is in the form of a π-series, which is a series of the form the sum from π
equals one to β of one over π to the π power. So letβs write out the condition
for convergence for a π-series. That is the π-series the sum from
π equals one to β of one over π to the π power is convergent if π is
greater than one and divergent if π is less than or equal to one. And so for this series, π equals
two-fifths which is less than one. So we can say that this series
diverges.

Determine whether the series the
sum from π equals one to β of one over the square root of π cubed converges
or diverges.

If we start by rewriting this sum,
using the fact that we can write the square root of π as π to the half power, we
can say that this sum is equivalent to the sum from π equals one to β of one
over π cubed raised to the half power. We can then use the fact that π to
the π₯ power, then raised to the π¦ power, equals π to the π₯ multiplied by π¦
power. So we can write our sum as the sum
from π equals one to β of one over π to the three over two power. And then we can recognise this to
be a π-series, which is a series of the form the sum from π equals one to β
of one over π to the π power.

So letβs write out the conditions
for convergence for a π-series. The π-series the sum from π
equals one to β of one over π to the π power is convergent if π is greater
than one and divergent if π is less than or equal to one. So for this series π is equal to
three over two; this is the same as 1.5, which is greater than one. So this series converges.

So letβs summarise the main points.
a π-series is a series of the form. the sum from π equals one to β of one
over π to the π power. And this is convergent if π is
greater than one and divergent if π is less than or equal to one. For π equals one, this series is
called the harmonic series. And this series is divergent.