### Video Transcript

Alternating Series Test

In this video, weβll learn how to
determine whether an alternating series is convergent using the alternating series
test. Weβll be looking at a variety of
examples of how we can use this convergence test. Letβs start by considering what an
alternating series is.

An alternating series is a series
which is of the form of the sum from π equal to one to β of negative one to the π
multiplied by π π. This is called an alternating
series, since the terms will switch between negative and positive values. This happens because we have a
negative one to the π. And so, when π is odd, negative
one to the π is negative one. And when π is even, negative one
to the π is simply one. Now, an alternating series may come
in a slightly different form. We could have the sum from π
equals one to β of negative one to the π plus one π π. And this would still be in
alternating series due to the negative one to the power of π plus one. However, in this case, when π is
odd, negative one to the π plus one is simply one. And when π is even, negative one
to the π plus one is negative one.

There are other types of series
which could be alternating series, which we should look out for. And these are Series, such as these
two here. If we look at the series on the
left, we have a cos of π π term. Letβs consider the first few values
of cos of π π. For π equals one, we have cos of
π, which is equal to negative one. For π equals two, we have cos of
two π, which is equal to one. For cos of three π, we have
negative one. And for cos of four π, we have
one. If we carried this on and on and
on, we would see the sign of one changing each time we increased π by one.

Now, when π is an odd number, we
have negative one. And when π is an even number, we
have positive one. Therefore, we can say that cos of
π π is equal to negative one to the π. And so, we can see that this first
series is in fact alternating. Similarly, we can consider sin of
two π plus one over two π. We have that sin of three π by two
is equal to negative one, sin of five π by two is equal to one, and sin of seven π
by two is equal to negative one. And therefore, here, we can also
see that sin of two π plus one over two π is equal to negative one to the π. Therefore, both of these series are
alternating series. And so, we must be careful when
identifying an alternating series, since some series may not immediately appear to
be an alternating series.

Now, thereβs one more condition
along with this, which must be satisfied for an alternating series. And that is that we require π π
to be greater than zero for all π. This is because if the sign of π
π is changing, then our series will no longer alternate between positive and
negative between terms. Also, if all of the π π terms are
less than zero, so theyβre negative, then we can simply take out a factor of
negative one and our series will satisfy these conditions to be an alternating
series.

Now that weβve considered what an
alternating series is, letβs have a look at how the alternating series test
works. So the alternating series test
tells us that for an alternating series, such as the one here, where π π is
greater than zero for all π, then if firstly, the limits as π tends to β of π π
is equal to zero and secondly, the sequence of π π is a decreasing sequence. Then, the series is a convergent
series. Letβs note that this test only
tells us if our series converges. If the test does not hold to be
true, this does not necessarily mean that our series diverges.

Letβs now look at an example of how
this test can be used.

Determine whether the series which
is the sum from π equals one to β of negative one to the π plus one multiplied by
five over π plus one minus five over π plus two converges or diverges.

Now, looking at our series, we can
see that it looks like an alternating series. This is because of the negative one
to the π plus one term. Now, an alternating series can be
of the form as shown here. However, weβre required to that π
π must be greater than zero. In our case, π π is equal to five
over π plus one minus five over π plus two. We can check whether this is
greater than zero by combining the two fractions. We multiply the fraction on the
left by π plus two over π plus two and multiply the fraction on the right by π
plus one over π plus one. After doing this, weβll have found
a common denominator of π plus one multiplied by π plus two. Then, we can combine the two
fractions. This will give five lots of π plus
two minus five lots of π plus one over π plus one multiplied by π plus two.

Now, we can multiply through the
numerator. And for the final step, we can
simplify this numerator. What weβre left with is that π π
is equal to five over π plus one multiplied by π plus two. Now, π can be any integer between
one and β. Therefore, π is always
positive. And if π is always positive, this
means that our π π must also always be positive. Now, we have confirmed that our
series is in fact an alternating Series. We can try and use the alternating
series test, which tells us that an alternating series converges if the limit as π
tends to β of π π is equal to zero and π π is a decreasing sequence.

Letβs start by finding the limit as
π tends to β of π π. We have that the limit as π tends
to β of π π is equal to the limit as π tends to β of five over π plus one
multiplied by π plus two. Now, all of the πs in our limit
are on the denominator of our fraction. And all of these πs are
positive. Therefore, as π gets larger and
larger and larger, the denominator of this fraction will also get larger. This means that the whole fraction
itself will tend towards zero. Hence, we can say that this limit
must be equal to zero. Therefore, weβve satisfied the
first condition of the alternating series test.

Next, we need to check whether the
sequence π π is decreasing. If the sequence is decreasing, then
π π must be greater than π π plus one, since each term will be smaller than the
previous one. We can rearrange this
inequality. And here, what we have to check is
that π π minus π π plus one is greater than zero. For π π, we have five over π
plus one times π plus two. Then, for π π plus one, we have
five over π plus two times π plus three. We can find the difference here by
multiplying the fraction on the left by π plus three over π plus three and the
fraction on the right by π plus one over π plus one.

After combining the two resulting
fractions, what weβre left with is five lots of π plus three minus five lots of π
plus one over π plus one times π plus two times π plus three. Simplifying this fraction, what
weβre left with is 10 over π plus one times π plus two times π plus three. Since π must be between one and β,
this tells us that π itself must be greater than or equal to one. For any π value, which is greater
than or equal to one, our fraction here, which is π π minus π π plus one, must
be greater than zero. This is because all of the terms in
this fraction will be positive. So we have satisfied the condition
required for π π to be a decreasing sequence. And therefore, the second condition
for the alternating series test is also satisfied.

From this, we can come to the
conclusion that the series which is the sum from π equals one to β of negative one
to the π plus one multiplied by five over π plus one minus five over π plus two
converges.

Next, letβs consider what happens
if we do not satisfy the conditions of the alternating series test.

Letβs suppose that we have the
alternating series, which is the sum from π equals one to β of negative one to the
π π π. And letβs say that weβve not
satisfied the first condition of the alternating series test. So there are limits as π tends to
β of π π is not equal to zero. Now, here, we can not directly say
that this series is divergent, since this is not stated within the alternating
series test. However, having this piece of
information about π π can be quite useful, especially when we consider the
divergence test.

We can recall that the divergent
test tells us that for some series, which is the sum from π equals one to β of π
π, if the limit as π tends to β of π π is nonzero, then the series diverges. Now, in the case of our alternating
series, we have that π π is equal to negative one to the π π π. Now, to check whether this
alternating series diverges, we need to find the limit of π π as π tends to
β. This is equivalent to the limit as
π tends to β of negative one to the π π π.

Using limit rules, we have that the
limit of a product is equal to the product of the limits. And so our limit here is equal to
the limit, as π tends to β of negative one to the π multiplied by the limit as π
tends to β of π π. Now, the second limit here, which
is the limit as π tends to β of π π, we have already defined to be nonzero.

Now, there are two cases which
could arise here. Maybe, this limit as π tends to β
of π π does not exist. If this is the case, then our limit
as π tends to β of negative one to the π π π does not exist either. This means that the limit as π
tends to β of π π does not exist. Therefore, it satisfies the
condition of the divergence test, since it will not be equal to zero. Therefore, we would be able to say
that our series diverges. Now, the other case is that our
limit as π tends to β of π π exists and is equal to some constant, which we can
call π, remembering that π is not equal to zero. If this is the case, then the limit
as π tends to β of negative one to the π times π π, which is also the limit of
π π will be equal to π multiplied by the limit as π tends to β of negative one
to the π.

Now, as π gets larger and larger
and larger, negative one to the π will keep flipping between negative one and
one. Therefore, it will not be
approaching any single value. And so, we can say that this limit
does not exist. Therefore, the limit as π tends to
β of π π does not exist either. And so, again, weβve satisfied the
condition of the divergence test. And we can say that our alternating
series diverges. Now this can be a pretty useful
method for testing divergence of alternating series, if our alternating series fails
the first part of the alternating series test.

Letβs now look at another
example.

Determine whether the series which
is the sum from π equals one to β of negative one to the π multiplied by π
squared plus three over five π squared minus one converges or diverges.

Now, our series looks like an
alternating series. An alternating series is a series
of the form shown here, where π π is greater than zero for all π. In our case, π π is equal to π
squared plus three over five π squared minus one. We can quite clearly see that both
the numerator and denominator of the fraction here will be positive for all π
greater than or equal to one. Therefore, it is true that π π is
greater than zero for all π. And so, our series is an
alternating series. We can, therefore, apply the
alternating series test.

This test tells us that a series
which of the form of the sum from π equals one to β of negative one to the π π
π, where π π is greater than zero for all π, converges. If firstly, the limit as π tends
to β of π π is equal to zero. And secondly, if π π is a
decreasing sequence. We can start by finding the limit
as π tends to β of π squared plus three over five π squared minus one. We can divide the numerator and
denominator by the highest power of π. So thatβs π squared. Weβre left with the limit as π
tends to β of one plus three over π squared over five minus one over π
squared.

Now, as π tends to β, any term
with π in the denominator. So thatβs three over π squared and
negative one over π squared will tend to zero. And weβll be left with this limit
being equal to one-fifth. Now, one-fifth is not equal to
zero. Therefore, we have failed to
satisfy the first condition of the alternating series test. And so, we can say that this series
does not converge by the alternating series test. However, we can now try and apply
the divergence test to this series.

The divergence test tells us that
if for the sum from π equals one to β of π π, the limit as π tends to β of π π
is nonzero, then the series diverges. Now, in our case, π π is equal to
negative one to the π multiplied by π squared plus three over five π squared
minus one. And this is also equal to negative
one to the π π π. Therefore, the limit as π tends to
β of π π is equal to the limit as π tends to β of negative one to the π π
π. And we can split this up using the
fact of the limit of the product is equal to the product of the limits. We are left with the limit as π
tends to β of negative one to the π times the limit as π tends to β of π π.

Now, weβve already found the limit
as π tends to β of π π and itβs equal to one-fifth. Now, we can consider the limit as
π tends to β of negative one to the π. Now, as π gets larger and larger
and larger, negative one to the π will simply be switching between negative one and
one. Therefore, we cannot say that this
limit tends to a particular value. And therefore, this limit does not
exist. Since this limit does not exist,
this means that the limit as π tends to β of π π also does not exist. Since it does not exist, this limit
is therefore not equal to zero. And this does in fact satisfy the
condition for the divergence test. Hence, we can come to the
conclusion that the sum from π equals one to β of negative one to the π times π
squared plus three over five π squared minus one diverges.

Next, letβs consider what happens
when our sequence of terms π π is not decreasing for all terms. If we have our alternating series,
which is the sum from π equals one to β of negative one to the π π π and we
notice that the sequence π π is not decreasing for all π. Now, in most cases, there is not a
lot we can do here. This series would probably fail the
alternating series test and we would not know anything else about it. However, in some cases there may be
something we can do.

Letβs suppose that π π is not
decreasing for all π, but instead is decreasing for π is greater than or equal to
some capital π, where capital π is simply just some number greater than one. If this is the case, we can rewrite
our alternating series. We can rewrite it as the sum from
π equals one to capital π minus one of negative one to the π π π plus the sum
from π is equal to capital π to β of negative one to the π π π. Now, we can say that this series on
the left is finite as long as each of the terms β so thatβs negative one to the π
π π β is itself finite too.

Now, if we consider the series on
the right, we have an alternating series for which our sequence of π π is
decreasing. So as long as the limit as π goes
to β of π π is equal to zero, then this alternating series satisfies the
alternating series test. Therefore, it converges. Now, since the sum of these two
series is equal to our original alternating series. If the series on the left is finite
and the series on the right converges, then this implies that the series from π
equals one to β of negative one to the π π π also itself converges. Now, this can be a useful skill to
keep in mind when using the alternating series test, especially when our alternating
series fails the second condition of the test.

Now that weβve seen a couple of
examples of the alternating series test and a variety of skills which we can use
when applying the alternating series test, letβs cover some key point of the
video.

Key Points

An alternating series is one of the
form of the sum from π equals one to β of negative one to the π π π or the sum
from π equals one to β of negative one to the π plus one π π, where π π is
greater than zero for all π. However, sometimes, negative one to
the π may appear differently, for example, as cos of π π.

The alternating series test: for an
alternating series of the form, which is the sum from π equals one to β of negative
one to the π π π. If firstly, the limit as π tends
to β of π π is equal to zero and secondly, if the sequence π π is a decreasing
sequence, then the series converges. The alternating series test only
tells us if a series is convergent, not if it is divergent. However, the divergence test can be
a useful follow-on if the first condition of the alternating series test fails. So this is if the limit as π tends
to β of π π is not equal to zero.

Sometimes, we may be required to
split our series if the sequence of π π is only decreasing for π is greater than
or equal to capital π. If this is the case, we can rewrite
our series as the sum from π equals one to capital π minus one of negative one to
the π π π plus the sum from π is equal to capital π to β of negative one to the
π π π. Now, the time which weβd do this
would be if the alternating series test failed on the second condition and our
sequence of π π is only decreasing for π is greater than or equal to capital
π. In this case, weβd split up our
series like this and potentially have a finite series plus a convergent infinite
series, which will tell us that our overall alternating series is convergent.