AP Calculus BC: Integral Test

When faced with an infinite series, determining whether its sum approaches a finite number (converges) or grows without bound (diverges) is a fundamental challenge. For series where the terms are generated by a function, the Integral Test provides a powerful and elegant bridge to the world of calculus. This test allows you to leverage your skills with improper integrals to draw a definitive conclusion about the behavior of an entire series, making it an indispensable tool for series that don't fit neatly into simpler tests like the geometric or p-series formats.

Understanding the Prerequisites: When Can You Use the Test?

You cannot apply the Integral Test to every series. It has three strict conditions that must be satisfied for the function $f(x)$ that generates the series terms $a_n=f(n)$. First, $f(x)$ must be positive for all $x\ge N$, where $N$ is the starting index of the series. You are only concerned with the "tail" of the series, so it's acceptable if the function is positive only from some point onward. Second, $f(x)$ must be continuous on the interval $[N,\infty )$. Breaks or holes in the function invalidate the necessary connection. Finally, and most crucially, $f(x)$ must be decreasing for $x\ge N$. Intuitively, this means each subsequent term in the series is smaller than the one before it.

Verifying that a function is decreasing typically involves computing its derivative: if $f^{\prime }(x)<0$ for $x\ge N$, the condition is met. For example, consider the series ${\sum }_{n=1}^{\infty }\frac{1}{n^2+1}$. Here, $f(x)=\frac{1}{x^2+1}$. It is clearly positive and continuous for $x\ge 1$. Its derivative is $f^{\prime }(x)=-\frac{2x}{(x^2+1{)}^2}$, which is negative for all $x>0$, confirming the function is decreasing. All three conditions are satisfied, so the Integral Test can be applied.

The Mechanics of the Test: Connecting Series to Integrals

The core statement of the Integral Test is precise: If $f(x)$ is positive, continuous, and decreasing for $x\ge N$, then the infinite series ${\sum }_{n=N}^{\infty }f(n)$ and the improper integral ${\int }_N^{\infty }f(x)dx$ either both converge or both diverge. The test does not tell you the sum of the series; it only tells you whether the sum is finite (convergent) or infinite (divergent).

The procedure is methodical:

1. Identify the function: Confirm that $a_n=f(n)$.
2. Verify the conditions: Check for positivity, continuity, and that $f(x)$ is decreasing on some interval $[N,\infty )$.
3. Set up and evaluate the improper integral: Compute ${\int }_N^{\infty }f(x)dx={\lim }_{t\to \infty }{\int }_N^{t}f(x)dx$.
4. Draw the conclusion: If the improper integral converges to a finite number, the series converges. If the improper integral diverges (to infinity or does not exist), the series diverges.

Let's apply this to our example, ${\sum }_{n=1}^{\infty }\frac{1}{n^2+1}$. We have verified the conditions. Now, evaluate the corresponding improper integral:

$${\int }_1^{\infty }\frac{1}{x^2+1}dx=\underset{t\to \infty }{\lim }{\int }_1^t\frac{1}{x^2+1}dx=\underset{t\to \infty }{\lim }{\left[\arctan (x)\right]}_1^t$$

$$=\underset{t\to \infty }{\lim }\left(\arctan (t)-\arctan (1)\right)=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}.$$

Since the integral converges to the finite value $\frac{\pi }{4}$, the Integral Test concludes that the series ${\sum }_{n=1}^{\infty }\frac{1}{n^2+1}$ also converges.

Worked Examples and Subtleties

A classic application is proving the convergence or divergence of the p-series, ${\sum }_{n=1}^{\infty }\frac{1}{n^p}$. For $p>0$, $f(x)=1/x^p$ is positive, continuous, and decreasing. The test integral is ${\int }_1^{\infty }{x}^{-p}dx$. This integral converges if $p>1$ and diverges if $p\le 1$, which is exactly the p-series rule you've learned. The Integral Test provides the calculus-based proof for that rule.

Now, examine the harmonic series, ${\sum }_{n=1}^{\infty }\frac{1}{n}$ (a p-series with $p=1$). The test integral is: $${\int }_1^{\infty }\frac{1}{x}dx=\underset{t\to \infty }{\lim }\ln |x|{|}_1^t=\underset{t\to \infty }{\lim }(\ln t-0)=\infty .$$ The improper integral diverges to infinity, so by the Integral Test, the harmonic series diverges. This integral offers a visual explanation: the area under the curve $f(x)=1/x$ from 1 to infinity is infinite, and the series terms $1/n$ represent the areas of left-endpoint rectangles that contain more area than the curve itself. If the integral's area is infinite, the sum of the rectangle areas (the series) must also be infinite.

A more subtle point is that the convergence of the integral does not mean the series sum equals the integral value. In our first example, the integral converged to $\frac{\pi }{4}\approx 0.7854$, but the series sum ${\sum }_{n=1}^{\infty }\frac{1}{n^2+1}$ is actually approximately 1.0767. The integral provides a boundary, not an equality.

Common Pitfalls

Applying the test without checking all conditions. This is the most frequent error. For the series ${\sum }_{n=1}^{\infty }\frac{{\sin }^2(n)}{n^2}$, $f(x)=\frac{{\sin }^2(x)}{x^2}$ is positive and continuous, but it is not monotonically decreasing—it oscillates as it decreases. The Integral Test does not apply directly. A different test, like the Comparison Test, would be needed.

Misapplying the test to alternating series. The Integral Test requires positive terms. For a series like ${\sum }_{n=1}^{\infty }\frac{(-1{)}^n}{n}$, you cannot use $f(x)=\frac{-1}{x}$ or $\frac{1}{x}$ directly because the function would not be positive. You must apply the test to the series of absolute values $\sum \frac{1}{n}$ to test for absolute convergence, which in this case diverges (as shown by the harmonic series).

Confusing the index of summation. The lower limit of the series must match the lower limit of the improper integral. For ${\sum }_{n=0}^{\infty }\frac{1}{n^2+1}$, you would evaluate ${\int }_0^{\infty }\frac{1}{x^2+1}dx$. Getting this wrong can lead to an incorrect conclusion about convergence, though it usually won't change the divergence/convergence outcome for well-behaved functions.

Incorrectly evaluating the improper integral. A procedural mistake in evaluating the limit, such as forgetting the limit process or making an integration error, will obviously lead to the wrong conclusion. Always write out the limit definition: ${\lim }_{t\to \infty }{\int }_N^{t}f(x)dx$.

Summary

• The Integral Test establishes that if $f(x)$ is positive, continuous, and decreasing for $x\ge N$, the series ${\sum }_{n=N}^{\infty }f(n)$ and the improper integral ${\int }_N^{\infty }f(x)dx$ share the same fate: they both converge or both diverge.
• The test is exceptionally useful for series whose terms are defined by a function that is easily integrable, and it provides the formal proof for the behavior of p-series.
• Its power is in its ability to transform a discrete summation problem into a continuous calculus problem you already know how to solve.
• Its major limitation is its strict set of conditions; always verify positivity, continuity, and that the function is decreasing before applying the test.
• Remember that a convergent integral provides a conclusive answer about series convergence but does not give the sum of the series. The value of the integral and the sum of the series are different numbers.