10.4 Integral Test for Convergence

Keywords

Integral Test 积分判别法 • Positive 正的

Continuous 连续的 • Decreasing 递减的 • Improper Integral 反常积分

The Bridge Between Sums and Integrals

The Integral Test states that if the terms of a series can be modeled by a well-behaved function, then the infinite series and the improper integral will share the exact same convergence behavior.

Theorem: The Integral Test (LIM-7.A.6)

Let \(f\) be a function that is positive, continuous, and decreasing for all \(x \geq 1\). Let \(a_n = f(n)\).

If the improper integral \(\int_{1}^{\infty} f(x) \, dx\) converges, then the series \(\sum\limits_{n=1}^{\infty} a_n\) converges.
If the improper integral \(\int_{1}^{\infty} f(x) \, dx\) diverges, then the series \(\sum\limits_{n=1}^{\infty} a_n\) diverges.

Visualizing the Test

Why does this work? We can view the terms of a series \(a_n\) as areas of rectangles with a width of \(1\).

In the graph below, the curve is \(f(x) = \frac{1}{x^2}\). The points represent the terms of the series \(\sum \frac{1}{n^2}\). Since the total area under the curve from \(1\) to \(\infty\) is finite, the sum of the terms (which are bounded by the curve) must also be finite.

The Three Golden Conditions

Before you can use the Integral Test on the AP Exam, you must state and verify that the corresponding function \(f(x)\) is:

Positive: The terms are above the x-axis.
Continuous: There are no vertical asymptotes or holes for \(x \geq 1\).
Decreasing: The derivatives are negative (\(f'(x) < 0\)) or the terms clearly shrink.

Example: Testing \(\sum_{n=1}^{\infty} \frac{1}{n^2}\)

Let \(f(x) = \frac{1}{x^2}\). This function is positive, continuous, and decreasing on \([1, \infty)\). Let's integrate:

\[ \int_{1}^{\infty} \frac{1}{x^2} \, dx = \lim_{t \to \infty} \left[ -\frac{1}{x} \right]_{1}^{t} = \lim_{t \to \infty} \left( -\frac{1}{t} + 1 \right) = 1 \]

Since the integral converges to \(1\), the series \(\sum_{n=1}^{\infty} \frac{1}{n^2}\) converges. (Note: This does not mean the sum of the series is \(1\). The integral only shares convergence, not the exact sum value).

Extension: Euler's Solution to the Basel Problem

We just proved that \(\sum_{n=1}^{\infty} \frac{1}{n^2}\) converges using the Integral Test. But what does it converge to? This was known as the Basel Problem. In 1734, the brilliant mathematician Leonhard Euler shocked the world by proving that the sum is exactly:

\[ \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} \approx 1.6449 \]

It is mind-bending that the infinite sum of squares of reciprocals yields a result involving \(\pi\), the circle constant!