In the previous section we saw how to relate a series to an
improper integral to determine the convergence of a series. While the integral test is a nice test, it
does force us to do improper integrals which aren’t always easy and in some
cases may be impossible to determine the convergence of.
For instance consider the following series.
In order to use the Integral Test we would have to integrate
and I’m not even sure if it’s possible to do this
integral. Nicely enough for us there is
another test that we can use on this series that will be much easier to use.
First, let’s note that the series terms are positive. As with the Integral Test that will be important
in this section. Next let’s note that we
must have since we are integrating on the interval . Likewise, regardless of the value of x we will always have .
So, if we drop the x from the
denominator the denominator will get smaller and hence the whole fraction will
get larger. So,
Now,
is a geometric
series and we know that since the series will converge and its value will
be,
Now, if we go back to our original series and write down the
partial sums we get,
Since all the terms are positive adding a new term will only
make the number larger and so the sequence of partial sums must be an
increasing sequence.
Then since,
and because the terms in these two sequences are positive we
can also say that,
Therefore, the sequence of partial sums is also a bounded
sequence. Then from the second section on sequences
we know that a monotonic and bounded sequence is also convergent.
So, the sequence of partial sums of our series is a
convergent sequence. This means that the
series itself,
is also convergent.
So, what did we do here?
We found a series whose terms were always larger than the original
series terms and this new series was also convergent. Then since the original series terms were
positive (very important) this meant that the original series was also
convergent.
To show that a series (with only positive terms) was
divergent we could go through a similar argument and find a new divergent
series whose terms are always smaller than the original series. In this case the original series would have
to take a value larger than the new series.
However, since the new series is divergent its value will be
infinite. This means that the original
series must also be infinite and hence divergent.
We can summarize all this in the following test.
Comparison Test
In other words, we have two series of positive terms and the
terms of one of the series is always larger than the terms of the other
series. Then if the larger series is
convergent the smaller series must also be convergent. Likewise, if the smaller series is divergent
then the larger series must also be divergent.
Note as well that in order to apply this test we need both series to start
at the same place.
A formal proof of
this test is at the end of this section.
Do not misuse this test.
Just because the smaller of the two series converges does not say
anything about the larger series. The
larger series may still diverge.
Likewise, just because we know that the larger of two series diverges we
can’t say that the smaller series will also diverge! Be very careful in using this test
Recall that we had a similar test for improper integrals back when we were
looking at integration techniques. So,
if you could use the comparison test for improper integrals you can use the
comparison test for series as they are pretty much the same idea.
Note as well that the requirement that and really only need to be true eventually. In other words, if a couple of the first
terms are negative or for a couple of the first few terms we’re
okay. As long as we eventually reach a
point where and for all sufficiently large n the test will work.
To see why this is true let’s suppose that the series start
at and that the conditions of the test are only
true for for and for at least one of the conditions is not
true. If we then look at (the same thing could be done for ) we get,
The first series is nothing more than a finite sum (no
matter how large N is) of finite
terms and so will be finite. So the
original series will be convergent/divergent only if the second infinite series
on the right is convergent/divergent and the test can be done on the second series
as it satisfies the conditions of the test.
Let’s take a look at some examples.
Example 1 Determine
if the following series is convergent or divergent.
Solution
Since the cosine term in the denominator doesn’t get too
large we can assume that the series terms will behave like,
which, as a series, will diverge. So, from this we can guess that the series
will probably diverge and so we’ll need to find a smaller series that will
also diverge.
Recall that from the comparison test with improper
integrals that we determined that we can make a fraction smaller by either
making the numerator smaller or the denominator larger. In this case the two terms in the
denominator are both positive. So, if we drop the cosine term we will in fact
be making the denominator larger since we will no longer be subtracting off a
positive quantity. Therefore,
Then, since
diverges (it’s harmonic or the pseries test) by the Comparison Test our original series must
also diverge.

Example 2 Determine
if the following series converges or diverges.
Solution
In this case the “+2” and the “+5” don’t really add
anything to the series and so the series terms should behave pretty much like
which will converge as a series. Therefore, we can guess that the original
series will converge and we will need to find a larger series which also
converges.
This means that we’ll either have to make the numerator
larger or the denominator smaller. We
can make the denominator smaller by dropping the “+5”. Doing this gives,
At this point, notice that we can’t drop the “+2” from the
numerator since this would make the term smaller and that’s not what we
want. However, this is actually the furthest that we need to go. Let’s
take a look at the following series.
As shown, we can write the series as a sum of two series
and both of these series are convergent by the pseries test. Therefore,
since each of these series are convergent we know that the sum,
is also a convergent series. Recall that the sum of two convergent
series will also be convergent.
Now, since the terms of this series are larger than the
terms of the original series we know that the original series must also be
convergent by the Comparison Test.

The comparison test is a nice test that allows us to do
problems that either we couldn’t have done with the integral test or at the
best would have been very difficult to do with the integral test. That doesn’t mean that it doesn’t have
problems of its own.
Consider the following series.
This is not much different from the first series that we
looked at. The original series converged
because the 3^{n} gets very
large very fast and will be significantly larger than the n. Therefore, the n doesn’t really affect the convergence
of the series in that case. The fact
that we are now subtracting the n off instead of adding the n on really
shouldn’t change the convergence. We can
say this because the 3^{n}
gets very large very fast and the fact that we’re subtracting n off won’t really change the size of
this term for all sufficiently large values of n.
So, we would expect this series to converge. However, the comparison test won’t work with
this series. To use the comparison test
on this series we would need to find a larger series that we could easily determine
the convergence of. In this case we
can’t do what we did with the original series.
If we drop the n we will make
the denominator larger (since the n
was subtracted off) and so the fraction will get smaller and just like when we
looked at the comparison test for improper integrals knowing that the smaller
of two series converges does not mean that the larger of the two will also
converge.
So, we will need something else to do help us determine the
convergence of this series. The
following variant of the comparison test will allow us to determine the
convergence of this series.
Limit Comparison Test
The proof of this
test is at the end of this section.
Note that it doesn’t really matter which series term is in
the numerator for this test, we could just have easily defined c as,
and we would get the same results. To see why this is, consider the following
two definitions.
Start with the first definition and rewrite it as follows,
then take the limit.
In other words, if c
is positive and finite then so is and if is positive and finite then so is c.
Likewise if then and if then . Both definitions will give the same results
from the test so don’t worry about which series terms should be in the
numerator and which should be in the denominator. Choose this to make the limit easy to
compute.
Also, this really is a comparison test in some ways. If c
is positive and finite this is saying that both of the series terms will behave
in generally the same fashion and so we can expect the series themselves to
also behave in a similar fashion. If or we can’t say this and so the test fails to
give any information.
The limit in this test will often be written as,
since often both terms will be fractions and this will make
the limit easier to deal with.
Let’s see how this test works.
Example 3 Determine
if the following series converges or diverges.
Solution
To use the limit comparison test we need to find a second
series that we can determine the convergence of easily and has what we assume
is the same convergence as the given series.
On top of that we will need to choose the new series in such a way as
to give us an easy limit to compute for c.
We’ve already guessed that this series converges and since
it’s vaguely geometric let’s use
as the second series.
We know that this series converges and there is a chance that since
both series have the 3^{n}
in it the limit won’t be too bad.
Here’s the limit.
Now, we’ll need to use L’Hospital’s Rule on the second
term in order to actually evaluate this limit.
So, c is
positive and finite so by the Comparison Test both series must converge since
converges.

Example 4 Determine
if the following series converges or diverges.
Solution
Fractions involving only polynomials or polynomials under
radicals will behave in the same way as the largest power of n will behave in the limit. So, the terms in this series should behave
as,
and as a series this will diverge by the pseries test. In fact, this would make a nice choice for
our second series in the limit comparison test so let’s use it.
So, c is
positive and finite and so both limits will diverge since
diverges.

Finally, to see why we need c to be positive and finite (i.e.
and ) consider the following two series.
The first diverges and the second converges.
Now compute each of the following limits.
In the first case the limit from the limit comparison test
yields and in the second case the limit yields . Clearly, both series do not have the same
convergence.
Note however, that just because we get or doesn’t mean that the series will have the
opposite convergence. To see this
consider the series,
Both of these series converge and here are the two possible
limits that the limit comparison test uses.
So, even though both series had the same convergence we got
both and .
The point of all of this is to remind us that if we get or from the limit comparison test we will know
that we have chosen the second series incorrectly and we’ll need to find a
different choice in order to get any information about the convergence of the
series.
We’ll close out this section with proofs of the two tests.
Proof of Comparison
Test
The test
statement did not specify where each series should start. We only need to require that they start at
the same place so to help with the proof we’ll assume that the series start
at . If the series don’t start at the proof can be redone in exactly the same
manner or you could use an index shift
to start the series at and then this proof will apply.
We’ll start off
with the partial sums of each series.
Let’s notice a
couple of nice facts about these two partial sums. First, because we know that,
So, both
partial sums form increasing sequences.
Also, because for all n
we know that we must have for all n.
With these
preliminary facts out of the way we can proceed with the proof of the test
itself.
Let’s start out
by assuming that is a convergent series. Since we know that,
However, we
also have established that for all n
and so for all n we also have,
Finally since is a convergent series it must have a finite
value and so the partial sums, are bounded above. Therefore, from the second section on sequences
we know that a monotonic and bounded sequence is also convergent and so is a convergent sequence and so is convergent.
Next, let’s
assume that is divergent. Because we then know that we must have as . However, we also know that for all n we have and therefore we also know that as .
So, is a divergent sequence and so is divergent.

Proof of Limit
Comparison Test