# What is the Integral Test for Convergence?

## Integral Convergence

Recall that one condition of the *Fundamental Theorem of Calculus*, is that the interval being investigated must be a closed interval. In mathematics and science, however, it is not uncommon to consider integrals that have one or both boundaries at *+/-* infinity. Integrals that are being considered on open intervals are called **improper integrals**, and to evaluate an improper integral, the limit of the integral is evaluated for convergence or divergence. **Integral convergence** is defined as an integral whose limit exists and is finite, and **integral divergence** is defined as an integral whose limit is either *+/-* {eq}\infty {/eq}, or nonexistent. When evaluating an integral with one boundary at infinity,

{eq}\int_a^{\infty} f(x) dx = lim_{A \rightarrow \infty} \int_a^{A} f(x) dx {/eq}.

If both of the boundaries of the integral are infinity, the integral must be broken up, such as

{eq}\int_{-\infty}^{\infty} f(x) dx = \int_{-\infty}^{a} f(x) dx + \int_{a}^{\infty} f(x) dx {/eq},

and both terms on the right must be evaluated for convergence or divergence. If either term is divergent, then so is then entire integral.

*Figure 1* is the graph of {eq}f(x) = \frac{1}{(0.5x-0.3)^2} {/eq}. Notice, on both sides of the peak, the function goes to 0. This is an example of a function that will have a convergent integral. *Figure 2*, on the other hand, is the graph of {eq}f(x) = \frac{1}{x} {/eq}. Notice on this graph that the function can be followed to *+/-* {eq}\infty {/eq}, without ever touching the x-axis. This is an example of a function whose integral is divergent.

## Integral Test for Convergence

What is the integral test? The concepts of integral convergence and divergence are extended to the study of mathematical series, in the form of the **integral test for convergence**. Recall that a series is a summation, and mathematical series, written as {eq}\sum_n^i f(n) {/eq}. This summation can be rewritten as an integral, {eq}\int_n^i f(x) dx {/eq}, and the connection between series and integrals makes it possible to use integral tests on series. If a series is written as {eq}\sum_n^{\infty} f(n) {/eq}, such that the summation is taken to infinity, it is possible to tell whether or not the series converges, by using the **integral test for series**. To use the integral test for convergence, the **integral test rules** state that the series must be continuous and decreasing. If these two conditions are met, testing for convergence can be computed as follows:

1) Write the series as an integral where f(x) is the integrand, and the limits of the series are the boundaries of the integral:

{eq}\sum_n^{\infty} f(n) = \int_n^{\infty} f(x) dx {/eq}.

2) Change the limits of integration on right side of 1):

{eq}\int_n^{\infty} f(x) dx = \int_n^{N} f(x) dx {/eq}.

3) Perform the integration:

{eq}\int_n^{N} f(x) dx = F(x) |_n^N {/eq}

4) Take the limit of the result of the integration:

{eq}lim_{N \rightarrow \infty} F(N) - F(n) {/eq}

If the limit is finite and it exists, the series converges. If the limit is infinity or the limit does not exist, DNE, the series diverges.

### Integral Test Conditions

The integral test for convergence can be used because taking the limit of a Riemann's sum for a function f(n), is analogous to taking the integral of the function f(x). This means, that if an integral is convergent, its associated series is too, but finding the convergence of the integral and series is not the same as finding the sum of the series. If an integral, or a series, is bounded and monotonically decreasing, or decreasing such that each successive value is less than or equal to the one before, it is always convergent. If these conditions are not met, then to find out whether or not the series and its associated integral are convergent or not, the **integral test conditions** must be met:

1) the series f(n) must be continuous on the interval {eq}[ n, \infty) {/eq}

2) the series f(n) must be monotonically decreasing

3) the starting value of the series must be the lower boundary of the integral

### Integral Test Examples

The following **integral test examples** show how to prove whether or not certain series are convergent or divergent.

Example 1: Prove that the harmonic series {eq}\sum_{n=1}^{\infty}\frac{1}{n} {/eq} is divergent.

1) Rewrite the series as an integral.

{eq}\sum_{n=1}^{\infty}\frac{1}{n} = \int_1^{\infty} \frac{1}{x} dx {/eq}

2) Adjust the boundaries of the integral.

{eq}\int_1^{\infty} \frac{1}{x} dx = \int_1^{N} \frac{1}{x} dx {/eq}

3) Find the antiderivative of the integrand.

{eq}\int_1^{N} \frac{1}{x} dx = ln(x) |_1^N {/eq}

{eq}ln(x) |_1^N = ln(N) - ln(1) = ln(N) {/eq}

4) Compute the limit as {eq}N \rightarrow \infty {/eq}.

{eq}lim_{N \rightarrow \infty} ln(N) = ln(\infty) = \infty {/eq}

This series diverges because the limit of its integral is infinity.

Example 2: Is the series {eq}\sum_{n=1}^{\infty} \frac{1}{(0.5n-0.3)^2} {/eq} convergent or divergent?

1) Rewrite the series as an integral.

{eq}\sum_{n=1}^{\infty} \frac{1}{(0.5n-0.3)^2} = \int_1^{\infty} \frac{1}{(0.5x-0.3)^2} {/eq}

2) Adjust the boundaries of the integral.

{eq}\int_1^{\infty} \frac{1}{(0.5x-0.3)^2} = \int_1^N \frac{1}{(0.5x-0.3)^2} {/eq}

3) Integrate the right side of 2), using u-substitution.

{eq}u = (0.5x-0.3)^2 {/eq}

{eq}du = 0.5x-0.3 dx = \sqrt(u) dx {/eq}

{eq}\int_1^N \frac{1}{(0.5x-0.3)^2} = \int_1^N \frac{1}{u \sqrt(u)} du {/eq}

{eq}\int_1^N \frac{1}{u \sqrt(u)} du = \int_1^N \frac{1}{u^{\frac{3}{2}}} du {/eq}

{eq}\int_1^N \frac{1}{u^{\frac{3}{2}}} du = \int_1^N u^{-\frac{3}{2}} du {/eq}

{eq}\int_1^N u^{-\frac{3}{2}} du = \frac{-2}{u^\frac{1}{2}}|_1^N {/eq}

{eq}\frac{-2}{u^\frac{1}{2}}|_1^N = -2( \frac{1}{N^\frac{1}{2}} - \frac{1}{1^\frac{1}{2}}) {/eq}

4) Take the limit of the right side of the result from 3).

{eq}lim_{N \rightarrow \infty} -2( \frac{1}{N^\frac{1}{2}} - \frac{1}{1^\frac{1}{2}}) = -2( \frac{1}{\infty^\frac{1}{2}} - \frac{1}{1^\frac{1}{2}}) {/eq}

{eq}-2( \frac{1}{\infty^\frac{1}{2}} - \frac{1}{1^\frac{1}{2}}) = \frac{2}{1^\frac{1}{2}} = 2 {/eq}

The limit of this integral is finite and real-valued so the series is convergent.

Example 3: Prove that, for all *p* series {eq}\sum_{n=1}^{\infty} \frac{1}{n^p} {/eq} where {eq}p > 1 {/eq}, the series is convergent.

1) Write the series as an integral.

{eq}\sum_{n=1}^{\infty} \frac{1}{n^p} = \int_1^{\infty} \frac{1}{x^p} dx {/eq}

2) Adjust the boundaries of the integral.

{eq}\int_1^{\infty} \frac{1}{x^p} dx = \int_1^N \frac{1}{x^p} dx {/eq}

3) Integrate the right side of 2).

{eq}\int_1^N \frac{1}{x^p} dx = \int_1^N x^{-p} dx {/eq}

{eq}\int_1^N x^{-p} dx = \frac{x^{-p+1}}{-p+1}|_1^N {/eq}

{eq}\int_1^N x^{-p} dx|_1^N = \frac{N^{-p+1}}{-p+1} - \frac{1^{-p+1}}{-p+1} {/eq}

4) Rearrange the final answer from 3) and take the limit.

{eq}lim_{N \rightarrow \infty} \frac{N^{1-p}}{1-p} - \frac{1}{1-p} = \frac{\infty^{1-p}}{1-p} - \frac{1}{1-p} {/eq}

{eq}\frac{\infty^{1-p}}{1-p} - \frac{1}{1-p} = - \frac{1}{1-p} {/eq}

If {eq}p > 1 {/eq} the limit of the integral is real-valued and finite, so any p-series where {eq}p > 1 {/eq}, is convergent.

## Lesson Summary

An **improper integral** is an integral that has one or both boundaries set at infinity, and, instead of looking for a solution to the integral, it is often enough to know whether or not the integral converges. **Integral convergence** occurs when the limit of the antiderivative of the integrand is finite and real-valued. **Integral divergence** occurs if either the limit of the integral does not exist, or if it is infinity.

The concepts of integral convergence and divergence can be extended to open series. Since it is possible to translate a mathematical series into an integral, a series that is continuous on the interval {eq}[n, \infty) {/eq}, can be written as an improper integral, and the limit of the integral can be taken. This method is called the **integral test for convergence**. If the integral converges, then so does its associated series. The integral test for convergence does not find the summation of the series, and the **integral test rules** state that to use the **integral test for series**, the series must not only be continuous, but it must also be monotonically decreasing. Once the **integral test conditions** are met, this is a useful tool for finding out whether or not a series converges.

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## Additional Examples of the Integral Test

In the following examples, students will solidify their knowledge of the integral test for series convergence by first verifying if the test can be used (by checking all of the conditions of the test) and then applying the test if it able to be used. When learning convergence tests for infinite series, a common mistake among learners is to use a test without verifying the conditions required in order for the test to be applicable. Using a test that is not valid leads to inaccurate results. After finishing these additional examples, students will have more confidence when using the integral test for series convergence and will have received necessary practice on verifying conditions prior to using tests.

## Problems

For all of the following examples, determine if the integral test for series convergence is applicable. If it can be used, then use the integral test for series convergence to determine if the series converges or diverges.

## Solutions

1) The integral test can be used because the corresponding function

Then, using the integral test, we have

The integral diverges and so the series also diverges.

2) The integral test cannot be used because the corresponding function (as well as the series) is not positive nor decreasing. The value of the numerator of the series alternates between 1 and -1, and so the integral test does not apply.

3) The integral test can be used because the corresponding function

which is negative when

The integral diverges and so the series also diverges.

#### How do you know if an integral converges?

To know if an integral converges, compute the antiderivative of the integrand, then take the limit of the result. If an integral converges, its limit will be finite and real-valued.

#### Can the integral test be used to test convergence?

Yes, in calculus, integral tests are used to test for convergence and divergence of integrals. These same integral tests can be used to test for the convergence or divergence of a series.

#### How do you test for integral convergence?

To test for integral convergence, first compute the antiderivative of the integrand. Next, take the limit of this antiderivative as the variable goes to infinity. If the result is finite and real-valued, the integral is convergent.

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