The last two tests that we looked at for series convergence
have required that all the terms in the series be positive. Of course there are many series out there that
have negative terms in them and so we now need to start looking at tests for
these kinds of series.
The test that we are going to look into in this section will
be a test for alternating series. An alternating series is any series, 
,
for which the series terms can be written in one of the following two forms.
There are many other ways to deal with the alternating sign,
but they can all be written as one of the two forms above. For instance,
There are of course many others, but they all follow the
same basic pattern of reducing to one of the first two forms given. If you should happen to run into a different
form than the first two, don’t worry about converting it to one of those forms,
just be aware that it can be and so the test from this section can be used.
Alternating Series
Test
A proof of this test
is at the end of the section.
There are a couple of things to note about this test. First, unlike the Integral Test and the
Comparison/Limit Comparison Test, this test will only tell us when a series
converges and not if a series will diverge.
Secondly, in the second condition all that we need to
require is that the series terms, will be eventually decreasing. It is possible for the first few terms of a
series to increase and still have the test be valid. All that is required is that eventually we
will have for all n
after some point.
To see why this is consider the following series,
Let’s suppose that for is not decreasing and that for is decreasing.
The series can then be written as,
The first series is a finite sum (no matter how large N is) of finite terms and so we can
compute its value and it will be finite.
The convergence of the series will depend solely on the convergence of
the second (infinite) series. If the
second series has a finite value then the sum of two finite values is also
finite and so the original series will converge to a finite value. On the other hand if the second series is
divergent either because its value is infinite or it doesn’t have a value then
adding a finite number onto this will not change that fact and so the original
series will be divergent.
The point of all this is that we don’t need to require that
the series terms be decreasing for all n. We only need to require that the series terms
will eventually be decreasing since we can always strip out the first few terms
that aren’t actually decreasing and look only at the terms that are actually
decreasing.
Note that, in practice, we don’t actually strip out the
terms that aren’t decreasing. All we do
is check that eventually the series terms are decreasing and then apply the
test.
Let’s work a couple of examples.
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Example 1 Determine
if the following series is convergent or divergent.
Solution
First, identify the bn
for the test.
Now, all that we need to do is run through the two
conditions in the test.
Both conditions are met and so by the Alternating Series
Test the series must converge.
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The series from the previous example is sometimes called the
Alternating Harmonic Series. Also, the could be or any other form of alternating sign and we’d
still call it an Alternating Harmonic Series.
In the previous example it was easy to see that the series
terms decreased since increasing n
only increased the denominator for the term and hence made the term
smaller. In general however we will need
to resort to Calculus I techniques to prove the series terms decrease. We’ll see an example of this in a bit.
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Example 2 Determine
if the following series is convergent or divergent.
Solution
First, identify the bn
for the test.
Let’s check the conditions.
So, the first condition isn’t met and so there is no
reason to check the second. Since this
condition isn’t met we’ll need to use another test to check convergence. In these cases where the first condition
isn’t met it is usually best to use the divergence test.
So, the divergence test requires us to compute the following
limit.
This limit can be somewhat tricky to evaluate. For a second let’s consider the
following,
Splitting this limit like this can’t be done because this
operation requires that both limits exist and while the second one does the
first clearly does not. However, it
does show us how we can at least convince ourselves that the overall limit
does not exist (even if it won’t be a direct proof of that fact).
So, let’s start with,
Now, the second
part of this clearly is going to 1 as while the first part just alternates between
1 and -1. So, as the terms are alternating between positive
and negative values that are getting closer and closer to 1 and -1
respectively.
In order for
limits to exist we know that the terms need to settle down to a single number
and since these clearly don’t this limit doesn’t exist and so by the
Divergence Test this series diverges.
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Example 3 Determine
if the following series is convergent or divergent.
Solution
Notice that in this case the exponent on the “-1” isn’t n or n+1. That won’t change how
the test works however so we won’t worry about that. In this case we have,
so let’s check the conditions.
The first is easy enough to check.
The second condition requires some work however. It is not immediately clear that these
terms will decrease. Increasing n to n+1 will increase both the numerator and the denominator. Increasing the numerator says the term
should also increase while increasing the denominator says that the term
should decrease. Since it’s not clear
which of these will win out we will need to resort to Calculus I techniques
to show that the terms decrease.
Let’s start with the following function and its
derivative.
Now, there are three critical points for this function, ,
,
and . The first is outside the bound of our
series so we won’t need to worry about that one. Using the test points,
and so we can see that the function in increasing on and decreasing on . Therefore, since we know as well that the bn are also increasing on and decreasing on .
The bn
are then eventually decreasing and so the second condition is met.
Both conditions are met and so by the Alternating Series
Test the series must be converging.
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As the previous example has shown, we sometimes need to do a
fair amount of work to show that the terms are decreasing. Do not just make the assumption that the
terms will be decreasing and let it go at that.
Let’s do one more example just to make a point.
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Example 4 Determine
if the following series is convergent or divergent.
Solution
The point of this problem is really just to acknowledge
that it is in fact an alternating series.
To see this we need to acknowledge that,
and so the series is really,
Checking the two condition gives,
The two conditions of the test are met and so by the
Alternating Series Test the series is convergent.
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It should be pointed out that the rewrite we did in previous
example only works because n is an
integer and because of the presence of the π. Without the π we couldn’t do this and if n wasn’t guaranteed to be an integer we
couldn’t do this.
Let’s close this section out with a proof of the Alternating
Series Test.
Proof of Alternating
Series Test
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Without loss of
generality we can assume that the series starts at . If not we could modify the proof below to
meet the new starting place or we could do an index shift to get the series to
start at .
First, notice
that because the terms of the sequence are decreasing for any two successive
terms we can say,
Now, let’s take
a look at the even partial sums.
So, is an increasing sequence.
Next, we can
also write the general term as,
Each of the
quantities in parenthesis are positive and by assumption we know that is also positive. So, this tells us that for all n.
We now know
that is an increasing sequence that is bounded
above and so we know
that it must also converge. So, let’s
assume that its limit is s or,
Next, we can
quickly determine the limit of the sequence of odd partial sums, ,
as follows,
So, we now know
that both and are convergent sequences and they both have
the same limit and so we also know
that is a convergent sequence with a limit of s.
This in turn tells us that is convergent.
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