Estimating the Value of
a Series
We have now spent quite a few sections determining the
convergence of a series, however, with the exception of geometric and
telescoping series, we have not talked about finding the value of a
series. This is usually a very difficult
thing to do and we still aren’t going to talk about how to find the value of a
series. What we will do is talk about
how to estimate the value of a series.
Often that is all that you need to know.
Before we get into how to estimate the value of a series
let’s remind ourselves how series convergence works. It doesn’t make any sense to talk about the
value of a series that doesn’t converge and so we will be assuming that the
series we’re working with converges.
Also, as we'll see the main method of estimating the value of series will
come out of this discussion.
So, let’s start with the series (the starting point is not important, but we
need a starting point to do the work) and let’s suppose that the series
converges to s. Recall that this means that if we get the
partial sums,
then they will form a convergent sequence and its limit is s.
In other words,
Now, just what does this mean for us? Well, since this limit converges it means
that we can make the partial sums, s_{n},
as close to s as we want simply by
taking n large enough. In other words, if we take n large enough then we can say that,
This is one method of estimating the value of a series. We can just take a partial sum and use that
as an estimation of the value of the series.
There are now two questions that we should ask about this.
First, how good is the estimation? If we don’t have an idea of how good the
estimation is then it really doesn’t do all that much for us as an estimation.
Secondly, is there any way to make the estimate better? Sometimes we can use this as a starting point
and make the estimation better. We won’t
always be able to do this, but if we can that will be nice.
So, let’s start with a general discussion about the
determining how good the estimation is.
Let’s first start with the full series and strip out the first n terms.
Note that we converted over to an index of i in order to make the notation
consistent with prior notation. Recall
that we can use any letter for the index and it won’t change the value.
Now, notice that the first series (the n terms that we’ve stripped out) is nothing more than the partial
sum s_{n}. The second series on the right (the one
starting at ) is called the remainder and denoted
by R_{n}. Finally let’s acknowledge that we also know
the value of the series since we are assuming it’s convergent. Taking this notation into account we can
rewrite (1)
as,
We can solve this for the remainder to get,
So, the remainder tells us the difference, or error, between
the exact value of the series and the value of the partial sum that we are
using as the estimation of the value of the series.
Of course we can’t get our hands on the actual value of the
remainder because we don’t have the actual value of the series. However, we can use some of the tests that
we’ve got for convergence to get a pretty good estimate of the remainder provided
we make some assumptions about the series.
Once we’ve got an estimate on the value of the remainder we’ll also have
an idea on just how good a job the partial sum does of estimating the actual
value of the series.
There are several tests that will allow us to get estimates
of the remainder. We’ll go through each
one separately.
Integral Test
Recall that in this case we will need to assume that the
series terms are all positive and will eventually be decreasing. We derived the integral test by using the
fact that the series could be thought of as an estimation of the area under the
curve of where . We can do something similar with the
remainder.
As we’ll soon see if we can get an upper and lower bound on the value of the remainder we can use these bounds to help us get upper and lower bounds on the value of the series. We can in turn use the upper and lower bounds on the series value to actually estimate the value of the series.
So, let’s first recall that the remainder is,
Now, if we start at ,
take rectangles of width 1 and use the left endpoint as the height of the
rectangle we can estimate the area under on the interval as shown in the sketch below.
We can see that the remainder, R_{n}, is the area estimation and it will overestimate the exact area. So, we have the
following inequality.
Next, we could also estimate the area by starting at ,
taking rectangles of width 1 again and then using the right endpoint as the
height of the rectangle. This will give
an estimation of the area under on the interval . This is shown in the following sketch.
Again, we can see that the remainder, R_{n}, is again this estimation and in this case it will
underestimate the area. This leads to
the following inequality,
Combining (2) and (3)
gives,
So, provided we can do these integrals we can get both an
upper and lower bound on the remainder.
This will in turn give us an upper bound and a lower bound on just how
good the partial sum, s_{n},
is as an estimation of the actual value of the series.
In this case we can also use these results to get a better
estimate for the actual value of the series as well.
First, we’ll start with the fact that
Now, if we use (2) we
get,
Likewise if we use (3) we
get,
Putting these two together gives us,
This gives an upper and a lower bound on the actual value of
the series. We could then use as an
estimate of the actual value of the series the average of the upper and lower
bound.
Let’s work an example with this.
Example 1 Using
to estimate the value of .
Solution
First, for comparison purposes, we’ll note that the actual
value of this series is known to be,
Using let’s first get the partial sum.
Note that this is “close” to the actual value in some
sense, but isn’t really all that close either.
Now, let’s compute the integrals. These are fairly simple integrals so we’ll
leave it to you to verify the values.
Plugging these into (4)
gives us,
Both the upper and lower bound are now very close to the
actual value and if we take the average of the two we get the following
estimate of the actual value.
That is pretty darn close to the actual value.

So, that is how we can use the Integral Test to estimate the
value of a series. Let’s move on to the
next test.
Comparison Test
In this case, unlike with the integral test, we may or may
not be able to get an idea of how good a particular partial sum will be as an
estimate of the exact value of the series.
Much of this will depend on how the comparison test is used.
First, let’s remind ourselves on how the comparison test
actually works. Given a series let’s assume that we’ve used the comparison
test to show that it’s convergent.
Therefore, we found a second series that converged and for all n.
What we want to do is determine how good of a job the
partial sum,
will do in estimating the actual value of the series . Again, we will use the remainder to do
this. Let’s actually write down the
remainder for both series.
Now, since we also know that
When using the comparison test it is often the case that the
b_{n} are fairly nice terms
and that we might actually be able to get an idea on the size of T_{n}. For instance, if our second series is a pseries we can use the results from
above to get an upper bound on T_{n}
as follows,
Also, if the second series is a geometric series then we
will be able to compute T_{n}
exactly.
If we are unable to get an idea of the size of T_{n} then using the comparison
test to help with estimates won’t do us much good.
Let’s take a look at an example.
Example 2 Using
to estimate the value of .
Solution
To do this we’ll first need to go through the comparison
test so we can get the second series.
So,
and
is a geometric series and converges because .
Now that we’ve gotten our second series let’s get the
estimate.
So, how good is it?
Well we know that,
will be an upper bound for the error between the actual
value and the estimate. Since our
second series is a geometric series we can compute this directly as follows.
The series on the left is in the standard form and so we
can compute that directly. The first
series on the right has a finite number of terms and so can be computed
exactly and the second series on the right is the one that we’d like to have
the value for. Doing the work gives,
So, according to this if we use
as an estimate of the actual value we will be off from the
exact value by no more than and that’s not too bad.
In this case it can be shown that
and so we can
see that the actual error in our estimation is,
Note that in this case the estimate of the error is
actually fairly close (and in fact exactly the same) as the actual
error. This will not always happen and
so we shouldn’t expect that to happen in all cases. The error estimate above is simply the upper
bound on the error and the actual error will often be less than this value.

Before moving on to the final part of this section let’s
again note that we will only be able to determine how good the estimate is
using the comparison test if we can easily get our hands on the remainder of
the second term. The reality is that we
won’t always be able to do this.
Alternating Series
Test
Both of the methods that we’ve looked at so far have
required the series to contain only positive terms. If we allow series to have negative terms in
it the process is usually more difficult.
However, with that said there is one case where it isn’t too bad. That is the case of an alternating series.
Once again we will start off with a convergent series which in this case happens to be an
alternating series that satisfies the conditions of the alternating series test, so we know that for all n. Also note that we could have any power on the
“1” we just used n for the sake of
convenience. We want to know how good of
an estimation of the actual series value will the partial sum, s_{n}, be. As with the prior cases we know that the
remainder, R_{n}, will be the
error in the estimation and so if we can get a handle on that we’ll know
approximately how good the estimation is.
From the proof of the
Alternating Series Test we can see that s
will lie between and for any n
and so,
Therefore,
We needed absolute value bars because we won’t know ahead of
time if the estimation is larger or smaller than the actual value and we know
that the b_{n}’s are positive.
Let’s take a look at an example.
Example 3 Using
to estimate the value of .
Solution
This is an alternating series and it does converge. In this case the exact value is known and
so for comparison purposes,
Now, the estimation is,
From the fact above we know that
So, our estimation will have an error of no more than
0.00390625. In this case the exact
value is known and so the actual error is,

In the previous example the estimation had only half the
estimated error. It will often be the
case that the actual error will be less than the estimated error. Remember that this is only an upper bound for
the actual error.
Ratio Test
This will be the final case that we’re going to look at for
estimating series values and we are going to have to put a couple of fairly
stringent restrictions on the series terms in order to do the work. One of the main restrictions we’re going to
make is to assume that the series terms are positive. We'll also be adding on another restriction in
a bit.
In this case we’ve used the ratio test to show that is convergent.
To do this we computed
and found that .
As with the previous cases we are going to use the
remainder, R_{n}, to
determine how good of an estimation of the actual value the partial sum, s_{n}, is.
Proof
Both parts will
need the following work so we’ll do it first.
We’ll start with the remainder.
Next we need to
do a little work on a couple of these terms.
Now use the
definition of to write this as,
Okay now let’s
do the proof.
For the first
part we are assuming that is decreasing and so we can estimate the
remainder as,
Finally, the
series here is a geometric series and because we know that it converges and we can compute
its value. So,
For the second
part we are assuming that is increasing and we know that,
and so we know
that for all n. The remainder can then be estimated as,
This is a
geometric series and since we are assuming that our original series converges
we also know that and so the geometric series above converges
and we can compute its value. So,

Note that there are some restrictions on the sequence and at least one of its terms in order to use
these formulas. If the restrictions
aren’t met then the formulas can’t be used.
Let’s take a look at an example of this.
Example 4 Using
to estimate the value of .
Solution
First, let’s use the ratio test to verify that this is a
convergent series.
So, it is convergent.
Now let’s get the estimate.
To determine an estimate on the remainder, and hence the
error, let’s first get the sequence .
The last rewrite was just to simplify some of the
computations a little. Now, notice
that,
Since this function is always decreasing and and so, this sequence is decreasing. Also note that .
Therefore we can use the first case from the fact above to get,
So, it looks like our estimate is probably quite
good. In this case the exact value is
known.
and so we can compute the actual error.
This is less than the upper bound, but unlike in the
previous example this actual error is quite close to the upper bound.

In the last two examples we’ve seen that the upper bound
computations on the error can sometimes be quite close to the actual error and
at other times they can be off by quite a bit.
There is usually no way of knowing ahead of time which it will be and
without the exact value in hand there will never be a way of determining which
it will be.
Notice that this method did require the series terms to be
positive, but that doesn’t mean that we can’t deal with ratio test series if
they have negative terms. Often series
that we used ratio test on are also alternating series and so if that is the
case we can always resort to the previous material to get an upper bound on the
error in the estimation, even if we didn’t use the alternating series test to
show convergence.
Note however that if the series does have negative terms,
but doesn’t happen to be an alternating series then we can’t use any of the
methods discussed in this section to get an upper bound on the error.