Integral Test for Convergence - AP Calculus BC Study Guide

The Core Idea: Integral Test for Convergence

The Integral Test provides a powerful method for determining the convergence or divergence of an infinite series by comparing it to an improper integral. The fundamental concept is that if the terms of a series, $a_n$, can be represented by a continuous, positive, and decreasing function, $f(x)$, where $a_n=f(n)$, then the infinite sum of the discrete terms of the series behaves in the same way as the infinite area under the continuous curve of the function.

Specifically, if the improper integral ${\int }_1^{\infty }f(x)dx$ converges to a finite value (meaning the area is finite), then the series ${\sum }_{n=1}^{\infty }{a}_n$ must also converge to a finite sum. Conversely, if the integral diverges (meaning the area is infinite), then the series must also diverge. This test establishes a direct link between the behavior of a series and its corresponding continuous function, allowing us to use the tools of integration to analyze series convergence.

Key Theorems

The Integral Test

If the function $f$ is continuous, positive, and decreasing for all $x\ge 1$, and if $a_n=f(n)$, then the series ${\sum }_{n=1}^{\infty }{a}_n$ and the improper integral ${\int }_1^{\infty }f(x)dx$ either both converge or both diverge.

Understanding the Conditions and Limitations

The application of the Integral Test is strictly dependent on satisfying three critical conditions for the associated function, $f(x)$, on the interval of integration. Failure to verify these conditions invalidates the test's conclusion.

1. Continuous: The function $f(x)$ must be continuous on the interval $[1,\infty )$. For a series starting at $n=N$, the function must be continuous on $[N,\infty )$.

2. Positive: The function $f(x)$ must be positive (i.e., $f(x)>0$) on the interval $[1,\infty )$.

3. Decreasing: The function $f(x)$ must be decreasing on the interval $[1,\infty )$. This is often the most involved condition to verify. A common method is to show that the derivative, $f^{\prime }(x)$, is negative on the interval.

It is crucial to understand a key limitation of this test: the Integral Test determines convergence or divergence only; it does not find the sum of the series. The value of the convergent integral is not equal to the sum of the convergent series. That is, if ${\int }_1^{\infty }f(x)dx=L$, it does not mean that ${\sum }_{n=1}^{\infty }{a}_n=L$.

Core Concepts & Rules

• Function-Series Connection: The Integral Test connects an infinite series $\sum a_n$ to a related improper integral $\int f(x)dx$ by defining $a_n=f(n)$.

• Required Conditions: To use the Integral Test, the function $f(x)$ must be shown to be continuous, positive, and decreasing for all $x$ in the interval of integration (e.g., $x\ge 1$).

• Shared Fate: If the three conditions are met, the series and the integral share the same convergence behavior.

  • If ${\int }_1^{\infty }f(x)dx$ converges, then ${\sum }_{n=1}^{\infty }{a}_n$ converges.

  • If ${\int }_1^{\infty }f(x)dx$ diverges, then ${\sum }_{n=1}^{\infty }{a}_n$ diverges.

• Value vs. Behavior: The test provides no information about the value of the sum of a convergent series. The value of the integral is not the sum of the series.

Step-by-Step Example 1: A Convergent p-Series

Determine if the series ${\sum }_{n=1}^{\infty }\frac{1}{n^2+1}$ converges or diverges.

Step 1: Define the related function.

Let $f(x)=\frac{1}{x^2+1}$. The terms of the series are given by $a_n=f(n)$.

Step 2: Verify the three conditions for $x\ge 1$.

• Continuous:$f(x)$ is a rational function whose denominator, $x^2+1$, is never zero. Therefore, $f(x)$ is continuous for all real numbers, including the interval $[1,\infty )$.

• Positive: For $x\ge 1$, the numerator (1) is positive and the denominator ($x^2+1$) is positive. Thus, $f(x)$ is positive on $[1,\infty )$.

• Decreasing: To check if the function is decreasing, we find its derivative.

  $$f^{\prime }(x)=\frac{d}{dx}(x^2+1{)}^{-1}=-1(x^2+1{)}^{-2}(2x)=\frac{-2x}{(x^2+1{)}^2}$$

  For $x\ge 1$, the numerator $(-2x)$ is negative and the denominator $((x^2+1{)}^2)$ is positive. Therefore, $f^{\prime }(x)<0$ for $x\ge 1$, which confirms that $f(x)$ is decreasing on this interval.

Step 3: Set up and evaluate the improper integral.

Since all three conditions are met, we can apply the Integral Test. We evaluate the improper integral ${\int }_1^{\infty }\frac{1}{x^2+1}dx$.

$${\int }_1^{\infty }\frac{1}{x^2+1}dx=\underset{b\to \infty }{\lim }{\int }_1^b\frac{1}{x^2+1}dx$$

The antiderivative of $\frac{1}{x^2+1}$ is $\arctan (x)$.

$$=\underset{b\to \infty }{\lim }[\arctan (x){]}_1^b$$

$$=\underset{b\to \infty }{\lim }(\arctan (b)-\arctan (1))$$

$$=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$$

Step 4: State the conclusion.

The improper integral ${\int }_1^{\infty }\frac{1}{x^2+1}dx$ converges to a finite value, $\frac{\pi }{4}$. Therefore, by the Integral Test, the series ${\sum }_{n=1}^{\infty }\frac{1}{n^2+1}$ also converges.

Step-by-Step Example 2: A Divergent Series

Determine if the series ${\sum }_{n=2}^{\infty }\frac{1}{n\ln (n)}$ converges or diverges.

Step 1: Define the related function.

Let $f(x)=\frac{1}{x\ln (x)}$. The terms of the series are given by $a_n=f(n)$. The series starts at $n=2$, so we will check conditions and integrate on the interval $[2,\infty )$.

Step 2: Verify the three conditions for $x\ge 2$.

• Continuous: The function $f(x)$ is undefined at $x=0$ and $x=1$. Since our interval is $[2,\infty )$, the function is continuous on this interval.

• Positive: For $x\ge 2$, both $x$ and $\ln (x)$ are positive. Therefore, $f(x)$ is positive on $[2,\infty )$.

• Decreasing: We find the derivative using the quotient rule.

  $$f^{\prime }(x)=\frac{(x\ln x)(0)-(1)\cdot \frac{d}{dx}(x\ln x)}{(x\ln x{)}^2}=\frac{-(\ln x+x\cdot \frac{1}{x})}{(x\ln x{)}^2}=\frac{-(\ln x+1)}{(x\ln x{)}^2}$$

  For $x\ge 2$, the numerator $-(\ln x+1)$ is negative and the denominator $(x\ln x{)}^2$ is positive. Thus, $f^{\prime }(x)<0$ for $x\ge 2$, which confirms that $f(x)$ is decreasing.

Step 3: Set up and evaluate the improper integral.

All conditions are met, so we evaluate ${\int }_2^{\infty }\frac{1}{x\ln (x)}dx$.

$${\int }_2^{\infty }\frac{1}{x\ln (x)}dx=\underset{b\to \infty }{\lim }{\int }_2^b\frac{1}{x\ln (x)}dx$$

We use u-substitution. Let $u=\ln (x)$, so $du=\frac{1}{x}dx$.

$$=\underset{b\to \infty }{\lim }{\int }_{\ln (2)}^{\ln (b)}\frac{1}{u}du$$

$$=\underset{b\to \infty }{\lim }[\ln |u|{]}_{\ln (2)}^{\ln (b)}$$

$$=\underset{b\to \infty }{\lim }(\ln (\ln (b))-\ln (\ln (2)))$$

As $b\to \infty$, $\ln (b)\to \infty$, and therefore $\ln (\ln (b))\to \infty$. The limit does not exist.

Step 4: State the conclusion.

The improper integral ${\int }_2^{\infty }\frac{1}{x\ln (x)}dx$diverges. Therefore, by the Integral Test, the series ${\sum }_{n=2}^{\infty }\frac{1}{n\ln (n)}$ also diverges.

Using Your Calculator

The Integral Test is an analytical test. A calculator cannot be used to formally prove that the conditions of the test (continuous, positive, decreasing) are met. The derivative must be found and analyzed by hand to justify the "decreasing" condition.

However, a calculator can be useful for checking your evaluation of the improper integral. To approximate the value of a convergent integral or to see if an integral appears to be diverging, you can use the numerical integration function.

For example, to check the result of Example 1, ${\int }_1^{\infty }\frac{1}{x^2+1}dx$:

1. Access the numerical integration function (e.g., fnInt( on a TI-84 or $\int ()$ on the calculator screen).

2. Enter the function, variable, lower limit, and a very large upper limit.

   • fnInt(1/(x^2+1), X, 1, 10000)

3. The calculator will return a value close to $\frac{\pi }{4}\approx 0.7854$. This suggests the integral converges, supporting your analytical conclusion.

For Example 2, calculating `fnInt(1/(X*ln(X)), X, 2, 10000)will yield a large number. Trying an even larger upper bound like $100000 will yield an even larger number, suggesting the integral is diverging. This is only for confirmation, not a formal proof.

AP Exam Quick Hit

Common Question Types

• Direct Application: You will be given a series and asked to determine if it converges or diverges. The prompt may or may not suggest using the Integral Test.

  • Example: "Does the series ${\sum }_{n=1}^{\infty }n{e}^{-n^2}$ converge or diverge? Justify your answer."

• Verifying Conditions: A multiple-choice question might ask for which of several series the Integral Test can be applied. This requires you to check the "continuous, positive, decreasing" conditions for each option.

  • Example: "For which of the following series can the Integral Test be used to determine convergence or divergence?"

    (A) $\sum \frac{\cos (\pi n)}{n}$

    (B) $\sum \frac{1}{n^2+\sin (n)}$

    (C) $\sum \frac{e^n}{e^n+1}$

    (D) $\sum \frac{\ln (n)}{n^2}$

    (You would need to check which related function is eventually positive and decreasing).

Common Mistakes

• Forgetting to Justify Conditions: The most common error is failing to explicitly state and verify that the related function $f(x)$ is continuous, positive, and decreasing. No credit is typically awarded for a conclusion based on the Integral Test if the conditions are not checked.

• Claiming the Sum Equals the Integral: A major conceptual error is stating that the sum of the series is equal to the value of the improper integral. The test only shows that they share the same convergence behavior.

• Errors in Integration: Mistakes in finding the antiderivative or evaluating the limit of the improper integral are common. Be especially careful with u-substitution and limits involving infinity.

• Incorrect Derivative for "Decreasing": When verifying the decreasing condition, students may calculate the derivative of $f(x)$ incorrectly, leading to a false conclusion about whether the function is decreasing.