### Video Transcript

Integral test for series

In this video, we will learn how to
use the integral test for series to determine whether a series containing
nonnegative terms is convergent or divergent. We will be looking at a variety of
examples of how the integral test can be used. Letβs start by covering what the
integral test is. The integral test for series tells
us that if we have some π of π₯ which is a continuous, positive, decreasing
function on the interval between π and β and that π of π is equal to π π, then, firstly, if the integral from
π to β of π of π₯ with respect to π₯ is convergent, so is the sum from π equals
π to β of π π. And secondly, if the integral from
π to β of π of π₯ with respect to π₯ is divergent, then so is the sum from π
equals one to β of π π.

Now remember, if an integral or sum
is convergent, then it is finite. And if it is divergent, then it is
infinite. So essentially, what weβre doing
with this test is using the convergence of an integral with an infinite bound to
find the convergence of a series. Now, when weβre finding our
function π of π₯, itβs very important that these three conditions are
satisfied. Now, the first two, continuous and
positive, are usually quite easy to spot. However, sometimes we may need to
do a bit of extra work to check that π of π₯ is decreasing on the interval between
π and β. We can have a look at why this test
works intuitively by considering the following graphs.

Letβs suppose we have the series
which is the sum from π equals π to β of π π such that the function π of π,
which is equal to π π, is continuous, positive, and decreasing on the interval
between π and β. We can sketch a graph of π of π₯,
and it may look something like this. We can label π, π plus one, π
plus two, and so on on our π₯-axis. Now, we have two possibilities for
π of π₯. Now, in one case, the integral from
π to β of π of π₯ with respect to π₯ is convergent. If this is the case, then we can
draw bars on our graph to represent the terms in our series. Now, these bars that weβve drawn
here have a width of one and a height of π of π₯. Therefore, the area of each of
these bars is simply π of π₯.

Now, for the first bar, π of π₯ is
equal to π of π. And we have that π of π is equal
to π π, and so the area of this bar is π π. Similarly, the area of the next bar
is π π plus one. And the area of the next bar is π
π plus two, and so on, and so on. Now, if we sum the area of each of
these bars together all the way up to β, then what weβll have is our series. But quite clearly, each of these
bars is underneath our curve, π of π₯. And the area under the curve is
represented by this integral. So, we can see that our series must
be smaller than our integral. Since our integral is convergent,
this implies that our series must also be convergent.

Now, we consider the other case,
which is, if our integral from π to β of π of π₯ with respect to π₯ is
divergent. If this is the case, then we can
draw bars on our graph to represent our series as follows. Here, weβve drawn the bars slightly
differently. However, since the bars still have
a width of one and a height of π of π₯, the areas of the bars are still the terms
in our series. And so, the sum of the area of
these bars will be equal to our series.

Now, this time we have that the
integral from π to β of π of π₯ with respect to π₯ is divergent. And we can see that the area of the
bars representing our series is at above the line π of π₯. And so, they have a greater area
than the integral from π to β of π of π₯ with respect to π₯. Since this integral is divergent,
this tells us that our series must also be divergent. Now that weβve covered how the
integral test for series works, letβs look at some examples.

Determine whether the series, which
is the sum from π equals zero to β of π to the negative π, converges or
diverges.

We can try to find the convergence
of this series using the integral test. The integral test tells us that if
we have a function, π of π₯, which is continuous, positive, and decreasing on the
interval between π and β and that π of π is equal to π π. Then, if the integral from π to β
of π of π₯ with respect to π₯ is convergent, so is the sum from π equals π to β
of π π. And if the integral from π to β of
π of π₯ with respect to π₯ is divergent, so is the sum from π equals π to β of π
π. Now, the series weβre trying to
find the convergence of is this sum from π equals zero to β of π to the negative
π. Therefore, π π is equal to π to
the power of negative π. Since π of π is equal to π π,
we obtain that π of π₯ is equal to π to the negative π₯. Since weβre taking the sum from π
is equal to zero, this means that π is equal to zero.

And we now need to check whether π
is a continuous, positive, and decreasing function on the interval between zero and
β. π to the negative π₯ can also be
written as one over π to the π₯. Now, π to the π₯ is positive for
any value of π₯. Therefore, one over π to the π₯ is
also positive. So, weβve satisfied the condition
that π of π₯ is positive on our interval. Now, one over π to the π₯ is
continuous since π to the π₯ cannot be equal to zero at any value. Therefore, weβve also satisfied
this condition. And finally, since π to the π₯ is
an increasing function for all π₯, this means that one over π to the π₯ is a
decreasing function for all π₯. And so, weβve satisfied the last
condition on π of π₯.

Weβre therefore able to use the
integral test. We need to find the convergence of
the integral from zero to β of π to the negative π₯ with respect to π₯. We can use the fact that the
integral from π to the ππ₯ with respect to π₯ is equal to one over π π to the
ππ₯ plus π. In our case, π is equal to
negative one, and one over negative one is simply negative one. So, when we integrate π to the
negative π₯, we get negative π to the negative π₯. And this is, of course, between our
two bounds which is zero and β.

Since our upper bound is infinite,
we need to find the limit as π₯ tends to β of negative π to the negative π₯. And we mustnβt forget to subtract
negative π to the negative zero. When trying to evaluate this limit,
we can consider what happens as π₯ gets larger and larger and larger. As π₯ gets bigger, negative π₯ gets
more and more negative. Therefore, π to the power of
negative π₯ gets closer and closer to zero.

And so, we can say that this limit
is equal to zero. Then, we have π to the power of
negative zero, which is simply π to the power of zero. Anything to the power of zero is
one. And the two negative signs in front
of the term cancel out with one another to give us a positive sign. From here, we can evaluate that the
integral from zero to β of π to the negative π₯ with respect to π₯ is equal to
one. Therefore, we found that our
integral is convergent. And so, by the integral test, we
can say that the sum from π equals zero to β of π to the negative π is
convergent.

Letβs now move on to another
example.

Use the integral test to determine
whether the series, which is the sum from π equals one to β of one over π,
converges or diverges.

Letβs start by recalling the
integral test. The integral test tells us that if
π is a continuous, positive, decreasing function on the interval between π and β
and that π of π is equal to π π. Then, if the integral from π to β
of π of π₯ with respect to π₯ is convergent, so is the sum from π equals π to β
of π π. And secondly, if the integral from
π to β of π of π₯ with respect to π₯ is divergent, so is the sum from π equals π
to β of π π. Now, in our case, our series is the
sum from π equals one to β of one over π. Therefore, π π is equal to one
over π. Using the fact that π of π is
equal to π π, we can say that π of π₯ is equal to one over π₯.

Now, we need to check that π of π₯
is a continuous, positive, and decreasing function on the interval between π and
β. Noting that since our sum goes from
π equals one, π must be equal to one. Letβs start by checking the
continuity of our function over this interval. The only discontinuity of π of π₯
occurs when π₯ is equal to zero. However, zero is not in our
interval between one and β. Therefore, this function must be
continuous between one and β. On this interval, π₯ is always
positive. Therefore, one over π₯ must also be
positive. And so, weβve satisfied that
condition too. Now, if π₯ starts at one and gets
larger and larger and larger, then one over π₯ will get smaller and smaller and
smaller. Therefore, we can say that this is
a decreasing function on our interval.

Now that weβve satisfied these
three conditions, weβre able to use the integral test. We need to work out whether the
integral from one to β of one over π₯ with respect to π₯ is convergent or
divergent. Now, we know that the differential
of the natural logarithm of π₯ with respect to π₯ is equal to one over π₯. Therefore, the natural logarithm of
π₯ is the antiderivative of one over π₯. Hence, we have that our integral is
equal to the natural logarithm of π₯ between one and β. When using the bound of β, we need
to take the limit as π₯ tends to β of the natural logarithm of π₯. Then, for our lower bound of one,
we simply subtract the natural logarithm of one.

Now, letβs evaluate this limit, the
natural logarithm as an increasing function. Therefore, as π₯ gets larger, the
natural logarithm of π₯ also gets larger. Hence, we can say that this limit
is also equal to β. We also have that the natural
logarithm of one is equal to zero. However, regardless of what
constant this term is, this will not affect our answer of β. This tells us that our integral
must be divergent. Hence, by the integral test, we can
say that the sum from π equals one to β of one over π is divergent.

Now, when weβre using the integral
test, itβs not actually required for π of π₯ to be decreasing for all π₯ in our
interval between π and β. All we require is for π₯ to be
decreasing for all π₯ greater than or equal to some constant π, where π is greater
than or equal to π. Weβll see how this can be done in
the next example.

Determine whether the series, which
is the sum from π equals one to β of the square root of the natural logarithm of π
over π, converges or diverges.

We can use the integral test to
test this convergence. The integral test tells us that if
we have a function π of π₯ which is continuous, positive, and decreasing on the
interval between π and β and that π of π is equal to π π. Then, firstly, if the integral from
π to β of π of π₯ with respect to π₯ is convergent, so is the sum from π equals
π to β of π π. And secondly, if the integral from
π equals π to β of π of π₯ with respect to π₯ is divergent, so is the sum from π
equals π to β of π π. Now, our series is the sum from π
equals one to β of the square root of the natural logarithm of π over π. And so, we can say that π π is
equal to the square root of the natural logarithm of π over π. Since π of π is equal to π π,
π of π₯ is equal to the square root of the natural logarithm of π₯ over π₯.

We also have that π is equal to
one. We now need to check whether π of
π₯ is continuous, positive, and decreasing on the interval between one and β. Firstly, for continuity, we need
the inside of the square root, so thatβs the natural logarithm of π₯, to be
nonnegative. If the natural logarithm of π₯ is
greater than or equal to zero, this means that π₯ must be greater than or equal to
one. Since our value of π is one, we
are looking at the interval between one and β. Therefore, π₯ is always greater
than or equal to one. And so, the numerator of our
function is continuous. The only other discontinuity may
occur when the denominator of our fraction is equal to zero. However, π₯ is always greater than
or equal to one, and so π₯ cannot be equal to zero. Therefore, our function is
continuous on the interval.

Next, we need to check whether our
function is positive over this interval. The square root function in the
numerator is always positive, since itβs a positive square root. And the denominator is always
positive, since π₯ is greater than or equal to one. Therefore, weβve satisfied this
condition. Now, we need to check that our
function is decreasing between one and β. Now, itβs quite difficult to check
this without performing a calculation. So, we can find the values of π₯
for which π of π₯ is decreasing by differentiating π with respect to π₯. This will give us π prime of
π₯. Now, π of π₯ is a quotient. Therefore, we can use the quotient
rule. The quotient rule tells us that the
differential of π’ over π£ is equal to π£ multiplied by the differential of π’ minus
π’ multiplied by the differential of π£ all over π£ squared.

Now, in our case, π’ is equal to
the square root of the natural logarithm of π₯ and π£ is equal to π₯. Finding dπ£ by dπ₯ is fairly
straightforward. We differentiate π₯ and simply get
one. Now, when we differentiate the
square root of the natural logarithm of π₯, we need to use the chain rule. Now, we may find it easier to
rewrite π’. And we can rewrite it as the
natural logarithm of π₯ to the power of one-half.

To find dπ’ by dπ₯, we multiply by
the power, decrease the power by one, and finally multiply by the differential of
the inside of the function. So, thatβs the differential of the
natural logarithm of π₯. And this is simply equal to one
over π₯. Now, dπ’ by dπ₯ and dπ£ by dπ₯ is
simply π’ prime and π£ prime. And these can be substituted into
the formula for the quotient rule. We obtain that π prime of π₯ is
equal to π₯ multiplied by one-half multiplied by one over π₯ times the natural
logarithm of π₯ to the power of negative one-half minus the square root of the
natural logarithm of π₯ all over π₯ squared.

Now, weβre looking for the values
of π₯ such that π is decreasing. So, that means that π prime of π₯
is negative. Now, the denominator of π prime of
π₯, thatβs π₯ squared, is always positive since the square is always positive. Therefore, the values of π₯ for
which π prime is negative occur when the numerator of the fraction is negative. So, this is when the square root of
the natural logarithm of π₯ is greater than π₯ multiplied by one-half multiplied by
one over π₯ multiplied by the natural logarithm of π₯ to the power of negative
one-half. Here, we can cancel out the one
over π₯ and the π₯.

We can also rewrite the natural
logarithm of π₯ to the power of negative one-half as one over the square root of the
natural logarithm of π₯. Therefore, the right-hand side of
our inequality becomes one over two times the square root of the natural logarithm
of π₯. We can multiply both sides by the
square root of the natural logarithm of π₯. We obtain that π of π₯ is
decreasing for values of π₯, such that the natural logarithm of π₯ is greater than
one-half. Using a calculator, we find that
this is values of π₯ which are greater than 1.648 and so on. Therefore, we found that our
function is not decreasing over the entirety of our interval.

However, we are able to rewrite our
series. We can take the first term out of
our series so that our series is equal to the square root of the natural logarithm
of one over one plus the sum from π equals two to β of the square root of the
natural logarithm of π over π. Since the natural logarithm of one
is equal to zero, this term on its own is simply zero. And so, this new sum that we found
is in fact equal to the original sum. However, we are summing from π is
equal to two. Therefore, our value of π changes
to two. Now, we have that π of π₯ is
decreasing for π₯ is greater than 1.648. Therefore, weβve satisfied the last
condition to use the integral test and weβre able to use the integral test. So, letβs do that now.

We need to find out whether the
integral from two to β of the square root of the natural logarithm of π₯ over π₯
with respect to π₯ is convergent or divergent. In order to solve this integral, we
can in fact use a π’ substitution. We can let π’ be equal to the
square root of the natural logarithm of π₯. Rearranging this, we have that π₯
is equal to π to the power of π’ squared. We can differentiate π₯ with
respect to π’ to find that dπ₯ by dπ’ is equal to two π’π to the π’ squared. An equivalent statement is dπ₯ is
equal to two π’π to the π’ squared dπ’. When π₯ is equal to two, π’ is
equal to the square root of the natural logarithm of two. And when π₯ is equal to β, π’ is
also equal to β. Weβre now ready to perform our
substitution.

We can start by substituting in the
bounds of our integral, which will be the square root of the natural logarithm of
two and β. Then, we have that the square root
of the natural algorithm of π₯ is π’. And π₯ is π to the π’ squared. Finally, we need to substitute in
dπ₯. Now, we have completed our π’
substitution. Weβre able to cancel out the π to
the π’ squared in the denominator with the other π to the π’ squared. And our integral becomes the
integral from the square root of the natural logarithm of two to β of two π’ squared
dπ’.

We can integrate this by raising
the power of π’ by one and dividing by the new power. So, now we have two-thirds π’ cubed
between our two bounds. Substituting in our bounds, we
obtain this. And even though the second term
here is constant, the limit as π’ tends to β of two-thirds π’ cubed is in fact
infinite. Therefore, we have that this
integral is divergent. So, by the integral test, our
series from π equals two to β diverges. Since this series is equal to the
one in the question, we have found our solution, which is that our series is
divergent.

Weβve now covered a variety of
examples. Letβs recap some key points of the
video.

Key points

The integral test tells us that if
π is a continuous, positive, decreasing function on the interval between π and β
and that π of π is equal to π π. Then, firstly, if the integral from
π to β of π of π₯ with respect to π₯ is convergent, so is the sum from π equals
π to β of π π. And secondly, if the integral from
π to β of π of π₯ with respect to π₯ is divergent, so is the sum from π equals π
to β of π π. Sometimes our function may not be
decreasing on the interval between π and β, but weβre still able to use the
integral test as long as π of π₯ is decreasing for all values of π₯ greater than or
equal to π, where π is a constant such that π is greater than or equal to π.