Certain infinite series can be studied using improper integrals.

If , then diverges.

However, there are still some divergent series that the divergence test does not pick out! We begin this section with such an example that shows how there is a connection between certain special types of series and improper integrals.

*harmonic series*and will be important in the coming material. Notice that this series isis not geometric. It isis not telescoping, and , so the divergence test isis not conclusive. So what should we do?

We have seen that we can graph a sequence as a collection of points in the plane.
We consider the **harmonic sequence** where , and write out the ordered
list that represents it. If we plot the harmonic sequence, it looks like this.

As it turns out, there is a nice way to visualize the sum too! One such way is to to make rectangles whose areas are equal to the terms in the sequence.

Note that the height of the -th rectangle is precisely and the width of all of the rectangles is , so the area of the -th rectangle is .

Now, in order to conclude whether converges, we must analyze . Note that has a nice visual interpretation as the sum of the areas of the first rectangles, but since we don’t have an explicit formula for we can try to establish that

- is bounded and monotonic, and hence exists.
- is unbounded so does not exist.

How can we establish this? The previous image might remind you of a Riemann Sum and for good reason. This technique lets us visually compare the sum of an infinite series to the value of an improper integral. For instance, if we add a plot of to our picture above

Notice that the sum on the righthand side is simply . Since is an improper integral, so we need to determine whether exists. Notice

This means that is not bounded and hence and must diverge.

Now, let’s take a step back and see what we really needed in the previous example.

- We needed to find a function for which the area under the curve over any particular interval was less than the area of the rectangle whose height is to establish a lower bound for each . Note that we can always do this if is eventually positive and decreasingincreasing since we may view each as the area of the rectangle that coincides with at its lefthandrighthand endpoint.
- We needed the function to be “eventually continuous” so the improper integral can be computed as the limit of a single definite integral.

By “eventually” above, we really mean that should be continuous, positive, and
decreasing on some interval for some ; it doesn’t need to happen right away, but it
should hold for *all* real large enough -values. This leads us to an interesting
observation.

Let be an eventually continuous, positive, and decreasing function with . If diverges, so does .

That’s a pretty good observation, but we can do even better.

and try the same idea of visualizing the series as an area.

**exactly**represents to the sum We again visually compare the sum of an infinite series to the value of an improper integral by adding a plot of to our picture above.

Note that we have a slight annoyance if we consider since has a vertical asymptote at . This is easily avoided if we instead consider on the interval . This requires that we consider the rectangles on that interval too. We update our picture.

Notice that the sum of the areas of the rectangles now is the series .

This is not an issue, because we know

The value of the lower index of summation does not affect whether an infinite series converges or diverges.

That is, if we can show converges, then must also converge. Now to determine whether converges, let . We must determine if exists. Since we expect to converge, we expect to converge. To show this, we should establish is bounded and monotonic, and hence exists is unbounded so does not exist .

Notice that since for all , and , the sequence is increasingdecreasing .

To show that it is also bounded, note that for any , we can observe from the picture that

In particular, since we shade more area if we increase , we have that . By a routine calculation of the latter improper integral, we can show

and we thus have

Hence is both bounded and monotonic, so exists, and converges. Since converges, will also converge since the index where we start the summation does not affect whether the series converges.

Now, let’s take a step back and see what we really needed in the this example.

- We needed to find a function for which the area under the curve over any
particular interval was
*greater*than the area of the rectangle whose height is to establish a lower bound for each . Note that we can always do this if is eventually positive and decreasingincreasing since we may view each as the area of the rectangle that coincides with at its lefthandrighthand endpoint. - We needed to establish that the sequence of partial sums is eventually increasing. This must happen if all of the are positivenegative .
- We needed the function to be “eventually continuous” so the improper integral can be computed as the limit of a single definite integral.

By “eventually” above, we really mean that should be continuous, positive, and
decreasing on some interval ; it doesn’t need to happen right away, but it
should hold for *all* real large enough -values. This leads us to an interesting
observation.

Let be an eventually continuous, positive, and decreasing function with . If converges, so does .

### The Integral Test

The observations from the previous examples give us a new convergence test called
the *integral test*:

**either both converge or both diverge**.

Notice that, after performing a substitution if necessary,

so and hence the improper integral convergesdiverges . Thus, convergesdiverges .

The next examples synthesizes some concepts we have seen thus far.

is also bounded since for all (i.e. no terms are larger than or smaller than ).

For , note that

Since , we have that for all , so is increasing (and hence monotonic). To determine if bounded, we can determine whether exists. Indeed, since is increasing, we have that , so if the limit exists, it serves as an upper bound for all of the terms in the sequence.

We can write out the limit in this case and find

Thus, all we have to do is determine if the infinite series above converges.

To do this, we can apply the integral test. Let . Notice that is positive, continuous, and decreasing on , and

Since the improper integral converges, the series converges. Hence, exists so is bounded.

### p-Series

A very important type of series for future sections is , where . We call a series that
can be brought into this form a -*series*. We want to determine for which values of
these series converge and diverge.

Notice that in our model examples, both series were -series.

- The harmonic series is a -series with . It convergesdiverges .
- The series is a -series with . It convergesdiverges .

Note that if , the exponent is negative so . If , the exponent is psoitive so . Thus, only converges when and thus converges if and only if .

This result is important enough to list as a theorem.