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integral test youtube playlist

WikiBooks - Integral Test

Integral Test Quick Notes

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

17Calculus Subjects Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

Differential Equations

Search Practice Problems

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.

memorize to learn

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

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.

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 [11min-23secs]

video by Dr Chris Tisdell

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 [13min-43secs]

video by Dr Chris Tisdell

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 [3min-26secs]

video by PatrickJMT

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

PatrickJMT - Integral Test for Series: Why It Works [14min-46secs]

video by PatrickJMT

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 [7min-45secs]

video by PatrickJMT

Practice

Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems

Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. So, Practice A01 (1) is probably the first basic practice problem, A02 (2) is probably the second basic practice problem, etc. Practice B01 is probably the first intermediate practice problem and so on.

GOT IT. THANKS!

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

Basic Problems

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

Problem Statement

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

Final Answer

The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2} } }\) converges by the p-series test or the integral test.

Problem Statement

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

Solution

This is a p-series with \( p=2 > 1\), so the series converges by the p-series test.
We could also have used the Integral Test, as follows.

\(\displaystyle{ \int_{1}^{\infty}{\frac{1}{x^2}dx} }\)

\(\displaystyle{ \lim_{b \to \infty}{\int_{1}^{b}{\frac{1}{x^2}dx}} }\)

\(\displaystyle{ \lim_{b \to \infty}{\int_{1}^{b}{x^{-2}dx}} }\)

\(\displaystyle{ \lim_{b \to \infty}{ \left[ -x^{-1} \right]_{1}^{b}} }\)

\(\displaystyle{ \lim_{b \to \infty}{-b^{-1} + 1^{-1}} }\)

\( 0 + 1 = 1 \)

Since the improper integral is finite, the series converges by the Integral Test.
Note: The value \(1\), from the integral is NOT necessarily what the series converges to. The significance of this number is only that it is finite.

Final Answer

The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2} } }\) converges by the p-series test or the integral test.

close solution

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n(\ln n)^2} } }\)

Final Answer

The series converges by the integral test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n(\ln n)^2} } }\)

Solution

Although it doesn't affect the final answer, his last step in the video should be \(\displaystyle{ 0 - \left( \frac{-1}{\ln 2} \right) = \frac{1}{\ln 2}}\). He writes \(\ln 2\) instead of \(1/(\ln 2)\). Since \(1/(\ln 2)\) is finite, the series converges by the integral test.

240 solution video

video by MIT OCW

Final Answer

The series converges by the integral test.

close solution

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{(n^2+1)^2} } }\)

Final Answer

The series converges by the integral test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{(n^2+1)^2} } }\)

Solution

241 solution video

video by Dr Chris Tisdell

Final Answer

The series converges by the integral test.

close solution

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ ne^{-n^2} } }\)

Final Answer

The series converges by the integral test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ ne^{-n^2} } }\)

Solution

242 solution video

video by PatrickJMT

Final Answer

The series converges by the integral test.

close solution

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{n^2+1} } }\)

Final Answer

The series diverges by the integral test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{n^2+1} } }\)

Solution

243 solution video

video by PatrickJMT

Final Answer

The series diverges by the integral test.

close solution

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2+1} } }\)

Final Answer

The series converges by the integral test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2+1} } }\)

Solution

248 solution video

video by Krista King Math

Final Answer

The series converges by the integral test.

close solution

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{\sqrt{n}} } }\)

Final Answer

The series diverges by the integral test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{\sqrt{n}} } }\)

Solution

249 solution video

video by Educator.com

Final Answer

The series diverges by the integral test.

close solution

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n} } }\)

Final Answer

The series diverges by the integral test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n} } }\)

Solution

250 solution video

video by MIP4U

Final Answer

The series diverges by the integral test.

close solution

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{1.1}} } }\)

Final Answer

The series converges by the integral test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{1.1}} } }\)

Solution

251 solution video

video by MIP4U

Final Answer

The series converges by the integral test.

close solution

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{5n-2} } }\)

Solution

1213 solution video

video by MIP4U

close solution

Intermediate Problems

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^2-4n+5}}}\)

Final Answer

The series converges by the integral test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^2-4n+5}}}\)

Solution

Although this problem would be more easily solved using one of the comparison tests, this is a great example using the integral test.

244 solution video

video by PatrickJMT

Final Answer

The series converges by the integral test.

close solution

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{3^n} } }\)

Final Answer

The series converges by the integral test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{3^n} } }\)

Solution

245 solution video

video by PatrickJMT

Final Answer

The series converges by the integral test.

close solution

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n[(\ln n)^2+4]} } }\)

Final Answer

The series converges by the integral test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n[(\ln n)^2+4]} } }\)

Solution

246 solution video

video by PatrickJMT

Final Answer

The series converges by the integral test.

close solution

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[\frac{4}{n\sqrt[3]{n}}+\frac{5}{n}\right] } }\)

Final Answer

The series diverges by the integral test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[\frac{4}{n\sqrt[3]{n}}+\frac{5}{n}\right] } }\)

Solution

247 solution video

video by PatrickJMT

Final Answer

The series diverges by the integral test.

close solution

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\ln(n)}{n^2} } }\)

Final Answer

The series converges by the integral test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\ln(n)}{n^2} } }\)

Solution

252 solution video

video by MIP4U

Final Answer

The series converges by the integral test.

close solution

\(\displaystyle{ \sum_{k=2}^{\infty}{ \frac{2}{k\ln k} } }\)

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{k=2}^{\infty}{ \frac{2}{k\ln k} } }\)

Final Answer

The series diverges by the integral test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{k=2}^{\infty}{ \frac{2}{k\ln k} } }\)

Solution

2443 solution video

video by Math Cabin

Final Answer

The series diverges by the integral test.

close solution
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