There is a nice result for approximating the remainder for series that converge by the integral test.

**either both converge or both diverge**.

The key to proving that a series converges by the integral test is to note that if all of the terms in are eventually positive, then will be eventually increasing.

When , and we have the assumptions for the integral test, the picture to keep in mind is below.

When the improper integral converges, it can be used to establish an upper bound for . This means that will be bounded and monotonic and thus have a limit, which we can determine without finding an explicit formula for ! From the picture, it should also be clear that the series and the improper integral do not have the same value since the series is represented by the sum of the areas of all of the rectanglesthe area under the curve whereas the improper integral is represented by the sum of the areas of all of the rectanglesthe area under the curve .

As such, we made the following important observation.

When the assumptions for the integral test are met, we can use the integral test to determine if a series converges, but we cannot ever use it to find the value to which the series converges!

What, then should we do? Thankfully, the integral test comes with a nice remainder result, which we will state then explore in the context of a familiar example.

### Remainders and the Integral Test

We also have bounds for the value of the infinite series , which are constructed from the error bounds.

The series can thus be approximated by any value ini the above interval.

Note that we actually have two results here.

**The first set of inequalities gives bounds for the error.**

- The inequality tells us a
*lower bound*for the error; this means we know that the error made using is at least the value of . That is, if we use to approximate : - The inequality gives us an
*upper bound*for the error; this means we know that the error we make in our approximation can be no*more*than the value of . That is, if we use to approximate :

**The second result gives us a sharper estimate for the value of the series
than the usual estimate using just .**

Since we have a positivity assumption on the terms of the series, note that this means that for every , will be an overestimateunderestimate . Since we have a result for the minimum error made, we can add this to the value of to obtain the smallest possible value of the series.

Before commenting further, let’s explore how this test works in the context of an example we have seen in previous sections.

What if we want to **approximate** this sum within an error of . How many
terms should we sum? Is it enough to sum the first ten terms? The first
hundred? We want to find a number where we can be sure that While we could
determine the smallest such value for when we had an explicit formula for , we
are not so fortunate here. However, since we want to use a computer to
obtain the actual approximation, we do not need the smallest number ; we
only need one for which we can be sure that will approximate to within
.

As usual, the error in approximating by will be .

Consider the following graph.

Note that the area of the first rectangle is (which is precisely ) , the area of the second rectangle is , and so on. Thus, the sum of the areas of all of these rectangles is exactly . Comparing this to the area under the curve for yields

In other words, the remainder of the series must be less than the given integral!

We can perform a similar construction to find a lower bound for the error.

and conclude

Putting these together yields

which is precisely the first inequality guaranteed by the theorem in the context of this example.

Now, suppose that we want to approximate the value of the series to within . One way we can do this is to approximate by and find by requiring that the maximum error made is no more than . Using the fact that , we find

It seems that we need to be at least . Using a computer to sum the first terms we find to five decimal places that

We have previously mentioned that the value of this series actually is known to be , so the above sum certainly approximates the series to within , but there’s a second part of the theorem that provides a better approximation.

Note that all of the terms in our series are positivenegative , so the the approximation we found above must be an overestimateunderestimate . Since we know that the minimum error in this approximation made is , and we can readily compute that

we know that the *smallest* possible value for the series is .

Similarly, the *largest* error made is

so the *largest* possible value for the series is . Hence, we conclude that

We may approximate the value of the series by any point in this interval.

Wait a minute! We wanted to approximate the series in the last example to within of its actual value, but the actual bounds for the error led us to find

We thus actually know the value to within (and can verify this since we are given the actual value of the series). What’s going on here?

Let’s return to the theorem and make an important observation explicit.

- The inequality gives us the
*minimum*possible error made. - The inequality gives us the
*maximum*possible error made.

Since we have a positivity assumption on the terms, note that this means that for every , will be an overestimateunderestimate . This leads to an important observation; if we want to approximate a convergent series to within of its actual value, we can require that the difference between the upper estimate for the series and the lower estimate be no more than . Note that for every ,

If the difference between the left and right sides of the inequality is less than , then the infinite series must be within of its minimum possible value (the lefthand side) and its maximum possible value (the righthand side).

Now, note that

In going from the penultimate to the last step, we are justified in our calculation because the integrals both converge. This in practice will greatly reduce the number of terms needed to approximate the value of a series.

- Choose (or be given) a desired precision , meaning, determine how closely you want to approximate the infinite series.
- Find the value for from setting . Call this value .
- Approximate by noting that
Choose any value in the above interval to approximate the value of the series.

To help gain some intuition and see how this process works, let’s see how this sharpens our estimate for in the last example.

We can now compute .

In many cases, technology will be useful in finding the value of , but we can do this one by hand.

We can use the quadratic formula to find (we ignore the negative root since must be positive). Using a calculator, we find , so we should set .

We can finish the estimate by noting that to four decimal places, we have the following.

- .
- .
- .

Thus, we have the following interval of values in which the exact value of the series must lie.

We can approximate the value of the series by choosing any value in this interval. We will again choose the midpoint, and declare .

Note that we only needed terms in the infinite series by using this sharper technique, whereas we needed terms if we only used the upper bound for the error to tell us how many terms we need to add to obtain the desired accuracy.

One type of series that plays a major role in our study is that of a convergent -series, for . Approximating values for these to even a moderate degree of accuracy can be computationally taxing for a computer if we only use the upper bound for the error. For example, suppose we want to compute to within of its exact value. The curious reader can verify the following.

- If the upper bound for the error is used to establish a value for so is within of the exact value of the series, then . It takes Maple (a powerful CAS) a little over minutes to compute this.
- If the method in the previous example is used to establish a value for , then . It takes Maple seconds to compute this.

If the same attempt is made for the series ,

- If the upper bound for the error is used to establish a value for so is within of the exact value of the series, then . Maple crashes trying to compute this.
- If the method in the previous example is used to establish a value for , then . It takes Maple seconds to compute this.

The point? Technology has limitations! A little bit of mathematical theory can often go a long way, even when computers are used to perform the calculations.

I. Show that this series converges.

To find the necessary antiderivative, we use use integration by parts.

Thus, we find

Now, we can evaluate the improper integral needed for the integral test.

By growth rates, we have that , so we find that convergesdiverges to .

II. Using the integral test (and a calculator or computer!), use and the resulting error bounds to approximate the value of .

- .
- Upper error bound: .
- Lower error bound: .

and so we can construct the interval of values in which the value of the series must lie.

We can now approximate the series by choosing a value in this interval. We will again choose the midpoint, and obtain that .

This equation is impossible to solve by hand, but we can use technology to establish that we have equality when . As usual, is a little too small, so we choose

To find the approximate value, we first find and the error bounds.

- .
- Upper error bound: .
- Lower error bound: .

We can now construct the interval of values in which the value of the series must lie.

We can now approximate the series by choosing a value in this interval. We will again choose the midpoint, and obtain that .

### Summary

When we can establish that a series converges, but we do not have an explicit formula for , we cannot find the exact value of the series. However, we can sometimes approximate it by considering the sequence of remainders. There are two important types of questions we have asked about remainders thus far.

- How bad is the error made when we approximate a convergent infinite series by its first several terms?
- How many terms should we specify if we want to know the value of a convergent series to obtain a desired precision?

In some cases, we use a formula or bounds for the remainder to answer both of these, but when the series can be approximated using the integral test, we have a much more efficient way to estimate. We attack the previous questions in the following manner.

- To approximate using and the error bounds, do the following.
- Compute .
- Compute the
*minimum*error from . - Compute the
*maximum*error from .

The approximate value for the series is then found from putting these together.

- To specify the value of a convergent series to within a desired precision of , do
the following.
- Find a value for from setting .
- Compute .
- Approximate by noting that
Choose a value in this interval as the approximate value of the series. Frequently, the midpoint provides a good choice.