10.4 Integral Test for Convergence

The integral test lets you decide whether an infinite series converges or diverges by comparing it to an improper integral. If $f(x)$ is positive, continuous, and decreasing on $[k,\infty)$ and $a_n=f(n)$, then the series $\sum a_n$ and the integral $\int_k^{\infty} f(x)\,\text{d}x$ either both converge or both diverge. For AP Calculus BC, state the test's conditions before using the integral to justify convergence or divergence.

How Does the Integral Test Work?

The integral test turns a positive-term series into an improper integral problem. Match the term $a_n$ to a function $f(x)$, check that $f$ is positive, continuous, and decreasing, then use the convergence or divergence of $\int_k^\infty f(x)\,dx$ to decide the series.

This is a BC-only topic, and the integral test is the one to reach for when the terms of a series match a function you can integrate and the nth term test gives no answer.

When the Integral Test Applies and What It Shows

• The integral test applies only when $f(x)$ is positive, continuous, and decreasing on the interval, with $a_n=f(n)$.
• If $\int_k^{\infty} f(x)\,\text{d}x$ converges to a finite number, the series converges. If the integral diverges, the series diverges.
• The integral test tells you whether a series converges, not what value it converges to.
• Evaluate the improper integral with a limit: $\lim_{b\to\infty}\int_k^{b} f(x)\,\text{d}x$.
• This test is the foundation for the p-series rule and explains why the harmonic series diverges.
• Always check the three conditions before applying the test, or your justification is incomplete.

Integral Test Theorem

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

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

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

and the partial sums are bounded by:

$\int_{k}^{\infty}\ f(x)\ \text{d}x\leq \sum_{n=k}^{\infty}a_n\leq a_k+\int_k^{\infty}\ f(x)\ \text{d}x$

For now, focus on the first two points. Here is what each piece means.

Breaking Down the Conditions

You need a positive, decreasing function. Positive means it stays above zero over the entire interval. Decreasing means the values keep getting smaller as $x$ grows. Consider the function $f(x)=\frac{1}{x\ln(x)}$ over the interval $[2,\infty)$:

This function meets the conditions. Now you just need a series where $a_n=f(n)$, which is easy to set up:

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

With the conditions checked, you can apply the test.

Applying the Integral Test

Start by integrating the function:

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

Use $u$-substitution, where $u=\ln x$ and $\text{d}u=\frac{1}{x}\ \text{d}x$:

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

Undoing the substitution gives:

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

Now evaluate over the bounds:

$$\lim_{b\to\infty}\left(\ln(\ln b)-\ln(\ln 2)\right)=\infty$$

Because this improper integral is unbounded, it is divergent. By the integral test, the series $\sum_{n=2}^{\infty}\frac{1}{n\ln(n)}$ is also divergent.

An integral is convergent if it evaluates to a finite number, like 42. It is divergent if it evaluates as $\pm\infty$ or does not exist.

Justifying Series Convergence With an Improper Integral

Problem Solving:

• Confirm the three conditions first: positive, continuous, and decreasing on the interval. State them so your justification is complete.
• Match the series term to a function by replacing $n$ with $x$, so $a_n=f(n)$.
• Rewrite the improper integral as a limit, $\lim_{b\to\infty}\int_k^{b} f(x)\,\text{d}x$, and evaluate.
• State your conclusion clearly: the integral converges or diverges, so the series does the same.

Common Trap: The integral test only tells you whether a series converges. It does not give the sum of the series. The value of the integral and the value of the series are usually different numbers, so do not report the integral's value as the sum.

Integral Test Practice

Now apply what you've learned.

Integral Test Problems

1. State the integral test and its conditions.

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

   $$\sum_{n=0}^{\infty}\frac{1}{1+n^2}$$

3. Use the integral test to determine whether the series converges or diverges.

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

Integral Test Solutions

1. You need a function that is positive, continuous, and decreasing over an interval, plus a series whose terms come from that function with $a_n=f(n)$. If the matching improper integral converges, the series converges; if the integral diverges, the series diverges.

2. The integral $\int_{0}^{\infty}\frac{1}{1+x^2} \ \text{d}x$ is a standard integral, giving $[\arctan(x)]_0^{\infty}$ which gives $\pi/2-0=\pi/2$. Since the integral is finite, the series is convergent.

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

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

   Undoing $u$-substitution gives:

   $$\lim_{b\to\infty}\left[\frac{\ln(x^4+1)}{4}\right]_0^b=\infty$$

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

Common Misconceptions

• Misconception: The integral test gives the sum of the series.
  Actually: It does not. It only tells you whether the series converges or diverges.
• Misconception: You can skip checking the conditions.
  Actually: You cannot. If $f(x)$ is not positive, continuous, and decreasing on the interval, the test does not apply and your conclusion is not justified.
• Misconception: A convergent integral and its series share the same value.
  Actually: They usually do not. The integral and the series are different totals; the test only links their convergence behavior.
• Misconception: The starting index has to be 1.
  Actually: It does not. You can start at any value $k$ where the conditions hold, as long as the integral uses the same lower bound.
• Misconception: The integral test works for series with negative or alternating terms.
  Actually: It does not. The function must stay positive, so save alternating series for the alternating series test.

Related AP Calculus Guides

• Unit 10 Review: Infinite Series and Sequences
• 10.1 Defining Convergent and Divergent Infinite Series
• 10.3 The nth Term Test for Divergence
• 10.5 Harmonic Series and p-Series
• 10.2 Working with Geometric Series
• 10.6 Comparison Tests for Convergence