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You CAN Ace Calculus

17calculus > infinite series > integral test

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WikiBooks - Integral Test

Integral Test

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

Quick Summary

 used to prove convergence yes used to prove divergence yes can be inconclusive yes $$a_n$$ must be positive $$a_n$$ must be 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

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 $$\displaystyle{ \int_{k}^{\infty}{f(x)~dx} }$$.

 Two Things To Watch 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.

Okay, let's watch some videos to see how this test works.

In this first video clip, he does a great job explaining the integral test. He uses the integral test to show the divergence of the p-series $$\sum{ 1/n }$$.

 Dr Chris Tisdell - Intro to series + the integral test

In this next video, the instructor explains the integral test in more detail by using it on the two series $$\sum{ 1/n }$$ and $$\sum{ 1/n^2 }$$ to show that one diverges and the other converges.

 Dr Chris Tisdell - Integral test for Series

Here is another good explanation of the integral test. He looks at the sum $$\displaystyle{ \sum_{n=1}^{\infty}{\frac{1}{n^p}} }$$.

 PatrickJMT - Integral Test - Basic Idea

Here is a great video giving an intuitive understanding on why this works.

 PatrickJMT - 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.

 PatrickJMT - Remainder Estimate for the Integral Test

Search 17Calculus

Practice Problems

Instructions - - Unless otherwise instructed, determine the convergence or divergence of the following series using the integral test, if possible.

 Level A - Basic

Practice A01

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2} } }$$

solution

Practice A02

$$\displaystyle{\sum_{n=2}^{\infty}{\frac{1}{n(\ln n)^2}}}$$

solution

Practice A03

$$\displaystyle{\sum_{n=1}^{\infty}{\frac{n}{(n^2+1)^2}}}$$

solution

Practice A04

$$\displaystyle{\sum_{n=1}^{\infty}{ne^{-n^2}}}$$

solution

Practice A05

$$\displaystyle{\sum_{n=1}^{\infty}{\frac{n}{n^2+1}}}$$

solution

Practice A06

$$\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^2+1}}}$$

solution

Practice A07

$$\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{\sqrt{n}}}}$$

solution

Practice A08

$$\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n}}}$$

solution

Practice A09

$$\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^{1.1}}}}$$

solution

Practice A10

$$\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{5n-2}}}$$

solution

 Level B - Intermediate

Practice B01

$$\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^2-4n+5}}}$$

solution

Practice B02

$$\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{3^n}}}$$

solution

Practice B03

$$\displaystyle{\sum_{n=2}^{\infty}{\frac{1}{n[(\ln n)^2+4]}}}$$

solution

Practice B04

$$\displaystyle{\sum_{n=1}^{\infty}{\left[\frac{4}{n\sqrt[3]{n}}+\frac{5}{n}\right]}}$$

$$\displaystyle{\sum_{n=1}^{\infty}{\frac{\ln(n)}{n^2}}}$$