In this section we are going to take a look at a test that
we can use to see if a series is absolutely convergent or not. Recall that if a series is absolutely
convergent then we will also know that it’s convergent and so we will often use
it to simply determine the convergence of a series.
Before proceeding with the test let’s do a quick reminder of
factorials. This test will be
particularly useful for series that contain factorials (and we will see some in
the applications) so let’s make sure we can deal with them before we run into
them in an example.
If n is an integer
such that then n factorial is defined as,
Let’s compute a couple real quick.
In the last computation above, notice that we could rewrite
the factorial in a couple of different ways.
For instance,
In general we can always “strip out” terms from a factorial
as follows.
We will need to do this on occasion so don’t forget about
it.
Also, when dealing with factorials we need to be very
careful with parenthesis. For instance, as we can see if we write each of the
following factorials out.
Again, we will run across factorials with parenthesis so
don’t drop them. This is often one of
the more common mistakes that students make when they first run across
factorials.
Okay, we are now ready for the test.
Ratio Test
A proof of this test is at
the end of the section.
Notice that in the case of the ratio test is pretty much worthless and we
would need to resort to a different test to determine the convergence of the
series.
Also, the absolute value bars in the definition of L are absolutely required. If they are not there it will be possible for
us to get the incorrect answer.
Let’s take a look at some examples.
Example 1 Determine
if the following series is convergent or divergent.
Solution
With this first example let’s be a little careful and make
sure that we have everything down correctly.
Here are the series terms a_{n}.
Recall that to compute a_{n+1}
all that we need to do is substitute n+1
for all the n’s in a_{n}.
Now, to define L
we will use,
since this will be a little easier when dealing with
fractions as we’ve got here. So,
So, and so by the Ratio Test the series
converges absolutely and hence will converge.

As seen in the previous example there is usually a lot of
canceling that will happen in these.
Make sure that you do this canceling.
If you don’t do this kind of canceling it can make the limit fairly
difficult.
Example 2 Determine
if the following series is convergent or divergent.
Solution
Now that we’ve worked one in detail we won’t go into quite
the detail with the rest of these.
Here is the limit.
In order to do this limit we will need to eliminate the
factorials. We simply can’t do the
limit with the factorials in it. To
eliminate the factorials we will recall from our discussion on factorials
above that we can always “strip out” terms from a factorial. If we do that with the numerator (in this
case because it’s the larger of the two) we get,
at which point we can cancel the n! for the numerator an denominator to get,
So, by the Ratio Test this series diverges.

Example 3 Determine
if the following series is convergent or divergent.
Solution
In this case be careful in dealing with the factorials.
So, by the Ratio Test this series converges absolutely and
so converges.

Example 4 Determine
if the following series is convergent or divergent.
Solution
Do not mistake this for a geometric series. The n
in the denominator means that this isn’t a geometric series. So, let’s compute the limit.
Therefore, by the Ratio Test this series is divergent.

In the previous example the absolute value bars were
required to get the correct answer. If
we hadn’t used them we would have gotten which would have implied a convergent series!
Now, let’s take a look at a couple of examples to see what
happens when we get . Recall that the ratio test will not tell us
anything about the convergence of these series.
In both of these examples we will first verify that we get and then use other tests to determine the
convergence.
Example 5 Determine
if the following series is convergent or divergent.
Solution
Let’s first get L.
So, as implied earlier we get which means the ratio test is no good for
determining the convergence of this series.
We will need to resort to another test for this series. This series is an alternating series and so
let’s check the two conditions from that test.
The two conditions are met and so by the Alternating
Series Test this series is convergent.
We’ll leave it to you to verify this series is also absolutely
convergent.

Example 6 Determine
if the following series is convergent or divergent.
Solution
Here’s the limit.
Again, the ratio test tells us nothing here. We can however, quickly use the divergence
test on this. In fact that probably
should have been our first choice on this one anyway.
By the Divergence Test this series is divergent.

So, as we saw in the previous two examples if we get from the ratio test the series can be either
convergent or divergent.
There is one more thing that we should note about the ratio
test before we move onto the next section.
The last series was a polynomial divided by a polynomial and we saw that
we got from the ratio test. This will always happen with rational
expression involving only polynomials or polynomials under radicals. So, in the future it isn’t even worth it to
try the ratio test on these kinds of problems since we now know that we will
get .
Also, in the second to last example we saw an example of an
alternating series in which the positive term was a rational expression
involving polynomials and again we will always get in these cases.
Let’s close the section out with a proof of the Ratio Test.
Proof of Ratio Test
First note that
we can assume without loss of generality that the series will start at as we’ve done for all our series test
proofs.
Let’s start off
the proof here by assuming that and we’ll need to show that is absolutely convergent. To do this let’s first note that because there is some number r such that .
Now, recall
that,
and because we
also have chosen r such that there is some N such that if we will have,
Next, consider
the following,
So, for we have . Just why is this important? Well we can now look at the following
series.
This is a
geometric series and because we in fact know that it is a convergent
series. Also because by the Comparison test the series
is
convergent. However since,
we know that is also convergent since the first term on
the right is a finite sum of finite terms and hence finite. Therefore is absolutely convergent.
Next, we need
to assume that and we’ll need to show that is divergent. Recalling that,
and because we know that there must be some N such that if we will have,
However, if for all then we know that,
because the
terms are getting larger and guaranteed to not be negative. This in turn means that,
Therefore, by
the Divergence Test is divergent.
Finally, we
need to assume that and show that we could get a series that has
any of the three possibilities. To do
this we just need a series for each case.
We’ll leave the details of checking to you but all three of the
following series have and each one exhibits one of the
possibilities.
