The last topic that we discussed in the previous section was
the harmonic series. In that discussion
we stated that the harmonic series was a divergent series. It is now time to prove that statement. This proof will also get us started on the
way to our next test for convergence that we’ll be looking at.
So, we will be trying to prove that the harmonic series,
diverges.
We’ll start this off by looking at an apparently unrelated
problem. Let’s start off by asking what
the area under on the interval . From the section on Improper Integrals we know that
this is,
and so we called this integral divergent (yes, that’s the
same term we’re using here with series….).
So, just how does that help us to prove that the harmonic
series diverges? Well, recall that we
can always estimate the area by breaking up the interval into segments and then
sketching in rectangles and using the sum of the area all of the rectangles as
an estimate of the actual area. Let’s do
that for this problem as well and see what we get.
We will break up the interval into subintervals of width 1
and we’ll take the function value at the left endpoint as the height of the
rectangle. The image below shows the
first few rectangles for this area.
So, the area under the curve is approximately,
Now note a couple of things about this approximation. First, each of the rectangles overestimates
the actual area and secondly the formula for the area is exactly the harmonic
series!
Putting these two facts together gives the following,
Notice that this tells us that we must have,
Since we can’t really be larger than infinity the harmonic
series must also be infinite in value.
In other words, the harmonic series is in fact divergent.
So, we’ve managed to relate a series to an improper integral
that we could compute and it turns out that the improper integral and the
series have exactly the same convergence.
Let’s see if this will also be true for a series that
converges. When discussing the Divergence Test we made the
claim that
converges. Let’s see
if we can do something similar to the above process to prove this.
We will try to relate this to the area under is on the interval . Again, from the Improper Integral section we
know that,
and so this integral converges.
We will once again try to estimate the area under this
curve. We will do this in an almost
identical manner as the previous part with the exception that instead of using the left end points for the height of our rectangles we will use the
right end points. Here is a sketch of
this case,
In this case the area estimation is,
This time, unlike the first case, the area will be an
underestimation of the actual area and the estimation is not quite the series
that we are working with. Notice however
that the only difference is that we’re missing the first term. This means we can do the following,
Or, putting all this together we see that,
With the harmonic series this was all that we needed to say
that the series was divergent. With this
series however, this isn’t quite enough.
For instance and if the series did have a value of then it would be divergent (when we want
convergent). So, let’s do a little more
work.
First, let’s notice that all the series terms are positive
(that’s important) and that the partial sums are,
Because the terms are all positive we know that the partial
sums must be an increasing sequence. In other words,
In s_{n+1}
we are adding a single positive term onto s_{n}
and so must get larger. Therefore, the
partial sums form an increasing (and hence monotonic) sequence.
Also note that, since the terms are all positive, we can
say,
and so the sequence of partial sums is a bounded sequence.
In the second
section on Sequences we gave a theorem that stated that a bounded and
monotonic sequence was guaranteed to be convergent. This means that the sequence of partial sums
is a convergent sequence. So, who cares
right? Well recall that this means that
the series must then also be convergent!
So, once again we were able to relate a series to an
improper integral (that we could compute) and the series and the integral had
the same convergence.
We went through a fair amount of work in both of these
examples to determine the convergence of the two series. Luckily for us we don’t need to do all this
work every time. The ideas in these two
examples can be summarized in the following test.
Integral Test
A formal proof of this
test can be found at the end of this section.
There are a couple of things to note about the integral
test. First, the lower limit on the
improper integral must be the same value that starts the series.
Second, the function does not actually need to be decreasing
and positive everywhere in the interval.
All that’s really required is that eventually the function is decreasing
and positive. In other words, it is okay
if the function (and hence series terms) increases or is negative for a while,
but eventually the function (series terms) must decrease and be positive for
all terms. To see why this is true let’s
suppose that the series terms increase and or are negative in the range and then decrease and are positive for . In this case the series can be written as,
Now, 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.
A similar argument can be made using the improper integral
as well.
The requirement in the test that the function/series be
decreasing and positive everywhere in the range is required for the proof. In practice however, we only need to make
sure that the function/series is eventually a decreasing and positive
function/series. Also note that when
computing the integral in the test we don’t actually need to strip out the
increasing/negative portion since the presence of a small range on which the
function is increasing/negative will not change the integral from convergent to
divergent or from divergent to convergent.
There is one more very important point that must be made
about this test. This test does NOT give
the value of a series. It will only give
the convergence/divergence of the series.
That’s it. No value. We can use the above series as a perfect
example of this. All that the test gave
us was that,
So, we got an upper bound on the value of the series, but
not an actual value for the series. In
fact, from this point on we will not be asking for the value of a series we
will only be asking whether a series converges or diverges. In a later section
we look at estimating values of series, but even in that section still won’t
actually be getting values of series.
Just for the sake of completeness the value of this series
is known.
Let’s work a couple of examples.
Example 1 Determine
if the following series is convergent or divergent.
Solution
In this case the function we’ll use is,
This function is clearly positive and if we make x larger the denominator will get
larger and so the function is also decreasing. Therefore, all we need to do is determine
the convergence of the following integral.
The integral is divergent and so the series is also
divergent by the Integral Test.

Example 2 Determine
if the following series is convergent or divergent.
Solution
The function that we’ll use in this example is,
This function is always positive on the interval that
we’re looking at. Now we need to check
that the function is decreasing. It is
not clear that this function will always be decreasing on the interval
given. We can use our Calculus I
knowledge to help us however. The
derivative of this function is,
This function has two critical points (which will tell us
where the derivative changes sign) at . Since we are starting at we can ignore the negative critical
point. Picking a couple of test points
we can see that the function is increasing on the interval and it is decreasing on . Therefore, eventually the function will be
decreasing and that’s all that’s required for us to use the Integral Test.
The integral is convergent and so the series must also be
convergent by the Integral Test.

We can use the Integral Test to get the following fact/test
for some series.
Fact ( The p series Test)
Sometimes the series in this fact are called pseries and so this fact is sometimes
called the pseries test. This fact follows directly from the Integral
Test and a similar fact we saw in
the Improper Integral section. This fact
says that the integral,
converges if and diverges if .
Using the pseries
test makes it very easy to determine the
convergence of some series.
Example 3 Determine
if the following series are convergent or divergent.
(a)
(b)
Solution
(a) In this
case and so by this fact the series is
convergent.
(b) For this
series and so the series is divergent by the fact.

The last thing that we’ll do in this section is give a quick
proof of the Integral Test. We’ve
essentially done the proof already at the beginning of the section when we were introducing the Integral Test, but let’s go through it formally for a general function.
Proof of Integral
Test
First, for the
sake of the proof we’ll be working with the series . The original test statement was for a
series that started at a general and while the proof can be done for that it
will be easier if we assume that the series starts at .
Another way of
dealing with the is we could do an index shift and start the series at and then do the Integral Test. Either way proving the test for will be sufficient.
Let’s start off
and estimate the area under the curve on the interval and we’ll underestimate the area by taking
rectangles of width one and whose height is the right endpoint. This gives the following figure.
Now, note that,
The approximate
area is then,
and we know
that this underestimates the actual area so,
Now, let’s
suppose that is convergent and so must have a finite value. Also, because is positive we know that,
This in turn
means that,
Our series
starts at so this isn’t quite what we need. However, that’s easy enough to deal with.
So, just what
has this told us? Well we now know
that the sequence of partial sums, are bounded above by M.
Next, because
the terms are positive we also know that,
and so the
sequence is also an increasing sequence. So, we now know that the sequence of
partial sums converges and hence our series is convergent.
So, the first
part of the test is proven. The second
part is somewhat easier. This time
let’s overestimate the area under the curve by using the left endpoints of
interval for the height of the rectangles as shown below.
In this case
the area is approximately,
Since we know
this overestimates the area we also then know that,
Now, suppose
that is divergent. In this case this means that as because . However, because as we also know that .
Therefore,
since we know that as we must have . This in turn tells us that as .
So, we now know
that the sequence of partial sums, ,
is a divergent sequence and so is a divergent series.

It is important to note before leaving this section that in
order to use the Integral Test the series terms MUST eventually be decreasing
and positive. If they are not then the
test doesn’t work. Also remember that
the test only determines the convergence of a series and does NOT give the
value of the series.