# Calculus/Integral Test
## Integral Test

The Integral Test is easy to use and is good to use when the ratio test and the comparison tests won't work and you are pretty sure that you can evaluate the integral. The idea of this test is to evaluate the improper integral $\displaystyle {\int _{k}^{\infty }{f(x)~dx}}$.
The integral test utilizes the fact that an integral is essentially an Riemann Sum, which is itself an infinite sum, over an infinite interval. This is useful because integration is relatively straight forward and familiar.

### Integral Test Definition

For a series $\sum _{n=1}^{\infty }{a_{n}}$ where we can find a positive, continuous and decreasing function $f$ for $n>k$ and $a_{n}=f(n)$, then we know that if
converges, the series also converges. Similarly when the integral diverges, the series also diverges.

### Integral Test Quick Notes

| used to prove convergence | yes |
| used to prove divergence | yes |
| can be inconclusive | yes |
* $a_{n}$ must be positive and decreasing
* requires that the integrand must be integrable (not always possible)
* requires the evaluation of infinite limits (after integration)
* if the result of the limit (after integration) does not exist (different than diverges), this test is inconclusive

## Two Things to Watch Out For

1. The value of k
First, you need to find a constant k such that the function satisfies all of these conditions for all $n>k$:
* continuous
* positive
* decreasing
One of the favorite tricks that teachers like to put on exams (which I fell for when I first took the class) is to tell you to use the Integral Test but then not give you $k$. Many books just show this integral with $k=1$, which is not always valid. So be careful.
How To Find $k$: The best way is to calculate the critical values of the function and then check that the derivative is negative to the right of the largest critical value. Then, if you have access to a graphing calculator, do a quick plot to check your answer. If everything looks good, choose $k$ to be greater than the largest critical value. Any value will do, so choose one that will be easy to use in the integration. There is no one value that will always work. It depends on the function.
2. The final value of the integration Secondly, if you get a finite value for the integral and determine that the series converges, the finite value you got from the integral is NOT what the series converges to. The number itself has no meaning in this context (ie. we don't use the value of the number to tell us anything about the series). The significance of it lies in whether it is finite or not. That's it. That's all the information you can get from that number. So do NOT assume that the series converges to that number.

## Intuitive Explanation On Why This Test Works

Let us look at why this test works. As an example, we'll use the harmonic series $\sum _{n=1}^{\infty }{\frac {1}{n}}$. The harmonic series is a well known series that actually happens to be divergent. If we want to approximate the integral, we can use rectangles like in a Riemann Sum:
Note that this right-handed method will always under approximate the integral (assuming that the function is decreasing on our selected interval). This implies that if the right-handed sum is equal to the actual infinite series, then the integral itself must be greater than the sum. This can help show convergence, because if the integral from the starting point to infinity is convergent, then by the comparison test the original function on that interval must also be convergent. And so we can see that the integral test is actually a "special case" of the comparison test.
But what about divergence? This case is also satisfied—if we use a left-handed approximation instead of a right handed one, we see that we again attain the original series, however there is an important difference:
The key difference, in this case, is that the integral becomes an under approximation for the series, and we can use the new "series" of the integral to show divergence with the comparison test.
This test is useful but is unfortunately only useful on functions that can be integrated and are decreasing in size. The latter may seem like a trivial and unnecessary addition, however consider how this test works; it relies on the fact that the integral of a function decreasing on an interval will always yield an under/over approximation of a series; if the function is not decreasing everywhere on the interval, the integral will not necessarily yield an under/over approximation every time.

## Integral Test Proof

Here are 4 consecutive videos showing a proof of the Integral Test. You do not need to watch these in order to understand and use the Integral Test but we include them here for those who are interested.
* Integral Test Proof - Part 1 of 4
* Integral Test Proof - Part 2 of 4
* Integral Test Proof - Part 3 of 4
* Integral Test Proof - Part 4 of 4

## Video Recommendations

If you want a complete lecture on this test, we recommend this video. Calculus 2 Lecture 9.3: Using the Integral Test for Convergence/Divergence of Series, P-Series
In this video clip [11min-23secs], he does a great job of explaining the integral test. He uses the integral test to show the divergence of the p-series $\textstyle {\sum {1/n}}$.
Intro to series + the integral test
In this video, the instructor explains the integral test in more detail by using it on the two series $\textstyle {\sum {1/n}}$ and $\textstyle {\sum {1/n^{2}}}$ to show that one diverges and the other converges. Integral test for Series
Here is another good explanation of the integral test. He looks at the sum $\displaystyle {\sum _{n=1}^{\infty }{\frac {1}{n^{p}}}}$.
Integral Test - Basic Idea
Here is a great video giving an intuitive understanding on why this works. Integral Test for Series: Why It Works
This last video discusses the remainder estimate for the integral test. Although not required to understand how to use the integral test, this video will help you understand more intuitively what is going on. Remainder Estimate for the Integral Test

## Examples

Use the integral test to determine if the following series are convergent or divergent.

### Example 1

$\sum _{x=1}^{\infty }{\frac {x^{2}+1}{x}}$
This series does not pass the first requirement that the series is decreasing over the desired interval; $\lim _{x\rightarrow \infty }{\frac {x^{2}+1}{x}}=\infty$. Applying the integral test would still show the series' convergence.

### Example 2

$\sum _{n=1}^{\infty }{\frac {1}{(n-3)^{2}+1}}$
However, this series is not decreasing everywhere on the interval. However, it has a relative maximum at $n=1$, after which it does decrease forever. And so we can write this series as $\sum _{n=1}^{2}{\frac {1}{(n-3)^{2}+1}}+\sum _{n=3}^{\infty }{\frac {1}{(n-3)^{2}+1}}$. Integrating the function yields the improper integral $\tan ^{-1}(n-3)$ from $3$ to infinity, which converges, and so the series converges too.

## Integral Test Practice Problems

Determine the convergence or divergence of these series using the integral test, if possible.

### Practice Problems with Written Solutions

1. $\displaystyle {\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}}$
$k\geq 1$
$k\geq 1$
$p=2>1$
${\begin{array}{rcl}\displaystyle {\int _{1}^{\infty }{{\frac {1}{x^{2}}}dx}}&=&\displaystyle {\lim _{b\to \infty }{\int _{1}^{b}{{\frac {1}{x^{2}}}dx}}}\\&=&\displaystyle {\lim _{b\to \infty }{\int _{1}^{b}{x^{-2}dx}}}\\&=&\displaystyle {\lim _{b\to \infty }{\left[-x^{-1}\right]_{1}^{b}}}\\&=&\displaystyle {\lim _{b\to \infty }{-b^{-1}+1^{-1}}}\\&=&0+1=1\end{array}}$
$1$
$p=2>1$
${\begin{array}{rcl}\displaystyle {\int _{1}^{\infty }{{\frac {1}{x^{2}}}dx}}&=&\displaystyle {\lim _{b\to \infty }{\int _{1}^{b}{{\frac {1}{x^{2}}}dx}}}\\&=&\displaystyle {\lim _{b\to \infty }{\int _{1}^{b}{x^{-2}dx}}}\\&=&\displaystyle {\lim _{b\to \infty }{\left[-x^{-1}\right]_{1}^{b}}}\\&=&\displaystyle {\lim _{b\to \infty }{-b^{-1}+1^{-1}}}\\&=&0+1=1\end{array}}$
$1$

### Practice Problems with Video Solutions

| 1 | $\sum _{n=1}^{\infty }{\frac {1}{3^{n}}}$ | answer convergesconverges | solution | 2 | $\sum _{n=2}^{\infty }{\frac {1}{n[(\ln n)^{2}+4]}}$ | answer convergesconverges | solution |