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10.4 Integral Test for Convergence

1 min readjune 7, 2020

Welcome to AP Calc 10.4! In this guide, you’ll learn how to apply another test to determine whether series are convergent or divergent. This test will rely heavily on your understanding of indefinite integrals from Unit 6.


∫ Integral Test Theorem

The integral test states that, for a positive, decreasing function f(x)f(x) over the interval [k,)[k,\infty) with a corresponding sequence an=f(x)a_n=f(x), then…

(1) if k f(x) dx\int_k^{\infty}\ f(x)\ \text{d}x converges, then n=kan\sum_{n=k}^{\infty}a_n also converges,

(2) if k f(x) dx\int_k^{\infty}\ f(x)\ \text{d}x diverges, then n=kan\sum_{n=k}^{\infty}a_n also diverges,

and…

ak+k+1 f(x) dxn=kank f(x) dxa_k+\int_{k+1}^{\infty}\ f(x)\ \text{d}x\leq \sum_{n=k}^{\infty}a_n\leq \int_k^{\infty}\ f(x)\ \text{d}x

For now, we can ignore the comparison portion and just focus on the first points! Let’s break down what everything means.

🧱 Breaking Down the Integral Test Theorem

First, we need a positive, decreasing function. This means it must be above zero over the entire interval, and decreasing, or getting smaller over time. Consider the function f(x)=1x ln (x)f(x)=\frac{1}{x\ \text{ln}\ (x)} over the interval [2,)[2,\infty):

Untitled

Image courtesy of Desmos

This function satisfies our initial condition. Now, we just need a series where an=f(x)a_n=f(x). This is pretty easy to set up!

n=21n ln(n)\sum_{n=2}^{\infty}\frac{1}{n\ \text{ln}(n)}

Now, we can apply our test!

✏️ Applying the Integral Theorem

To apply our test, we first have to take the integral of our function:

21x ln (x) dx\int_2^{\infty} \frac{1}{x\ \text{ln}\ (x)}\ \text{d}x

To solve this integral, we need to apply uu-substitution, where u=ln xu=\text{ln}\ x and du=1x dx\text{d}u=\frac{1}{x}\ \text{d}x:

1u du=ln(u)\int \frac{1}{u}\ \text{d}u=\text{ln} (u)

Undoing uu-substitution, we find that

21x ln (x) dx=ln(ln(x))+C\int_2^{\infty} \frac{1}{x\ \text{ln}\ (x)}\ \text{d}x=\text{ln}(|\text{ln}(x)|)+C

Finally, we evaluate over our bounds:

ln(ln())ln(ln(2))=\text{ln}(|\text{ln}(\infty)|)-\text{ln}(|\text{ln}(2)|)=\infty

Because this integral evaluates as \infty, we say it is divergent. Based on our theorem, this means that the series n=21n ln(n)\sum_{n=2}^{\infty}\frac{1}{n\ \text{ln}(n)} is also divergent.


📝 Integral Test Theorem Practice

Now it’s your turn to apply what you’ve learned!

❓Integral Test Theorem Problems

  1. State the integral test and its conditions.

  2. Use the integral test to determine whether the series diverges.

    n=011+n2\sum_{n=0}^{\infty}\frac{1}{1+n^2}
  3. Use the integral test to determine whether the series diverges.

    n=0n3n4+1\sum_{n=0}^{\infty}\frac{n^3}{n^4+1}

💡 Integral Test Theorem Solutions

  1. Review the beginning of the guide for the full theorem! You need a function that is decreasing and positive over an interval and a series using that function.

  2. The integral 011+x2 dx\int_{0}^{\infty}\frac{1}{1+x^2} \ \text{d}x is a standard integral, giving [arctan(x)]0[\text{arctan}(x)]_0^{\infty} which gives π/20=π/2\pi/2-0=\pi/2, which means that the series is convergent.

  3. The integral 0x3x4+1 dx\int_{0}^{\infty}\frac{x^3}{x^4+1}\ \text{d}x is evaluated using uu-substitution, where u=x4+1u=x^4+1 and du=4x3 dx\text{d}u=4x^3\ \text{d}x:

    141u du=ln(u)4\frac{1}{4}\int\frac{1}{u}\ \text{d}u=\frac{\text{ln}(u)}{4}

    Undoing uu-substitution, we get:

    [ln(x4+1)4]0=0=\Bigg[\frac{\text{ln}(x^4+1)}{4}\Bigg]_0^\infty=\infty-0=\infty

    Because the integral is not finite, we say it diverges. Therefore, our series diverges.


    💫 Closing

    Excellent job! Be sure to practice this type of problem and make sure you know the conditions for using this test. Once you’ve got it down, the integral test will be a breeze! 💯

Key Terms to Review (2)

Diverges

: When a series or sequence diverges, it means that the terms of the series or sequence do not approach a finite value as the number of terms increases. In other words, the sum or limit of the terms does not exist.

Integral Test

: The integral test is a method used to determine the convergence or divergence of an infinite series by comparing it to the convergence or divergence of an improper integral.

10.4 Integral Test for Convergence

1 min readjune 7, 2020

Welcome to AP Calc 10.4! In this guide, you’ll learn how to apply another test to determine whether series are convergent or divergent. This test will rely heavily on your understanding of indefinite integrals from Unit 6.


∫ Integral Test Theorem

The integral test states that, for a positive, decreasing function f(x)f(x) over the interval [k,)[k,\infty) with a corresponding sequence an=f(x)a_n=f(x), then…

(1) if k f(x) dx\int_k^{\infty}\ f(x)\ \text{d}x converges, then n=kan\sum_{n=k}^{\infty}a_n also converges,

(2) if k f(x) dx\int_k^{\infty}\ f(x)\ \text{d}x diverges, then n=kan\sum_{n=k}^{\infty}a_n also diverges,

and…

ak+k+1 f(x) dxn=kank f(x) dxa_k+\int_{k+1}^{\infty}\ f(x)\ \text{d}x\leq \sum_{n=k}^{\infty}a_n\leq \int_k^{\infty}\ f(x)\ \text{d}x

For now, we can ignore the comparison portion and just focus on the first points! Let’s break down what everything means.

🧱 Breaking Down the Integral Test Theorem

First, we need a positive, decreasing function. This means it must be above zero over the entire interval, and decreasing, or getting smaller over time. Consider the function f(x)=1x ln (x)f(x)=\frac{1}{x\ \text{ln}\ (x)} over the interval [2,)[2,\infty):

Untitled

Image courtesy of Desmos

This function satisfies our initial condition. Now, we just need a series where an=f(x)a_n=f(x). This is pretty easy to set up!

n=21n ln(n)\sum_{n=2}^{\infty}\frac{1}{n\ \text{ln}(n)}

Now, we can apply our test!

✏️ Applying the Integral Theorem

To apply our test, we first have to take the integral of our function:

21x ln (x) dx\int_2^{\infty} \frac{1}{x\ \text{ln}\ (x)}\ \text{d}x

To solve this integral, we need to apply uu-substitution, where u=ln xu=\text{ln}\ x and du=1x dx\text{d}u=\frac{1}{x}\ \text{d}x:

1u du=ln(u)\int \frac{1}{u}\ \text{d}u=\text{ln} (u)

Undoing uu-substitution, we find that

21x ln (x) dx=ln(ln(x))+C\int_2^{\infty} \frac{1}{x\ \text{ln}\ (x)}\ \text{d}x=\text{ln}(|\text{ln}(x)|)+C

Finally, we evaluate over our bounds:

ln(ln())ln(ln(2))=\text{ln}(|\text{ln}(\infty)|)-\text{ln}(|\text{ln}(2)|)=\infty

Because this integral evaluates as \infty, we say it is divergent. Based on our theorem, this means that the series n=21n ln(n)\sum_{n=2}^{\infty}\frac{1}{n\ \text{ln}(n)} is also divergent.


📝 Integral Test Theorem Practice

Now it’s your turn to apply what you’ve learned!

❓Integral Test Theorem Problems

  1. State the integral test and its conditions.

  2. Use the integral test to determine whether the series diverges.

    n=011+n2\sum_{n=0}^{\infty}\frac{1}{1+n^2}
  3. Use the integral test to determine whether the series diverges.

    n=0n3n4+1\sum_{n=0}^{\infty}\frac{n^3}{n^4+1}

💡 Integral Test Theorem Solutions

  1. Review the beginning of the guide for the full theorem! You need a function that is decreasing and positive over an interval and a series using that function.

  2. The integral 011+x2 dx\int_{0}^{\infty}\frac{1}{1+x^2} \ \text{d}x is a standard integral, giving [arctan(x)]0[\text{arctan}(x)]_0^{\infty} which gives π/20=π/2\pi/2-0=\pi/2, which means that the series is convergent.

  3. The integral 0x3x4+1 dx\int_{0}^{\infty}\frac{x^3}{x^4+1}\ \text{d}x is evaluated using uu-substitution, where u=x4+1u=x^4+1 and du=4x3 dx\text{d}u=4x^3\ \text{d}x:

    141u du=ln(u)4\frac{1}{4}\int\frac{1}{u}\ \text{d}u=\frac{\text{ln}(u)}{4}

    Undoing uu-substitution, we get:

    [ln(x4+1)4]0=0=\Bigg[\frac{\text{ln}(x^4+1)}{4}\Bigg]_0^\infty=\infty-0=\infty

    Because the integral is not finite, we say it diverges. Therefore, our series diverges.


    💫 Closing

    Excellent job! Be sure to practice this type of problem and make sure you know the conditions for using this test. Once you’ve got it down, the integral test will be a breeze! 💯

Key Terms to Review (2)

Diverges

: When a series or sequence diverges, it means that the terms of the series or sequence do not approach a finite value as the number of terms increases. In other words, the sum or limit of the terms does not exist.

Integral Test

: The integral test is a method used to determine the convergence or divergence of an infinite series by comparing it to the convergence or divergence of an improper integral.


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© 2024 Fiveable Inc. All rights reserved.

AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.