Understanding the Integral Test for Series Convergence

The Integral Test is used to determine the convergence or divergence of an infinite series by comparing it to an improper integral. If you have a series (sum a_n), where (a_n = f(n)) and (f(x)) is a continuous, positive, and decreasing function for (geq 1), you can apply the Integral Test. The test compares the series to the improper integral of (f(x)) over the interval \([1, \infty)\): (int_1^infty f(x), dx) Check out online math resources for more practice.

• If the integral converges, the series also converges.
• If the integral diverges, the series also diverges.

The test is particularly useful for series where the terms resemble well-known functions that are easy to integrate, such as exponential or polynomial terms. However, if the function doesn’t meet the criteria of being positive, continuous, and decreasing, the test cannot be applied.

Here are two more examples demonstrating the Integral Test:

1. Converging Example:

Series: (sum_{n=1}^{infty} frac{e^{-n}}{n})

Function: (f(x) = frac{e^{-x}}{x})

Apply the Integral Test: Evaluate the integral: (int_1^infty frac{e^{-x}}{x}, dx)

This integral converges because the function (frac{e^{-x}}{x}) decays rapidly for large (x). Therefore, the series (sum_{n=1}^{infty} frac{e^{-n}}{n}) converges.

2. Diverging Example:

Series: (sum_{n=1}^{infty} frac{1}{n ln(n)})

Function: (f(x) = frac{1}{x ln(x)})

Apply the Integral Test: Evaluate the integral: (int_2^infty frac{1}{x ln(x)}, dx)

This is a known divergent integral (it diverges logarithmically). Therefore, the series (sum_{n=1}^{infty} frac{1}{n ln(n)}) diverges.

Frequently Asked Questions

Calculus for Beginners The Ultimate Step by Step Guide to Acing Calculus

Download

$27.99 Original price was: $27.99.$17.99Current price is: $17.99.

What is the integral test for convergence?

The integral test says that if f(x) is positive, continuous and decreasing for x ≥ 1, and aₙ = f(n), then the series ∑aₙ and the improper integral ∫₁^∞ f(x) dx either both converge or both diverge. It lets you use an integral, which is often easier to evaluate, to decide the behavior of a series.

How do you use the integral test for convergence?

First, write the terms as a function, such as aₙ = 1/n² becoming f(x) = 1/x². Next, check that f is positive, continuous and decreasing for x ≥ 1. Then evaluate the improper integral: ∫₁^∞ 1/x² dx = 1, which is finite, so ∑ 1/n² converges. If the integral had been infinite, the series would diverge.

What are the conditions for the integral test?

The integral test needs three conditions on f(x), where aₙ = f(n): f must be positive, continuous and decreasing, at least for all x beyond some starting value N. If any condition fails, the test cannot be used. For example, a function that oscillates, like f(x) = sin²(x)/x², is not decreasing, so you would need a comparison test instead.

Can the integral test prove divergence?

Yes. If the conditions hold and the improper integral diverges, the series diverges too. The harmonic series is the standard example: ∫₁^∞ 1/x dx = lim (ln b) as b → ∞, which grows without bound, so ∑ 1/n diverges. This is one of the clearest ways to show that terms going to 0 is not enough for convergence.

Does the integral test prove absolute convergence?

Since the integral test only applies to series with positive terms, any convergence it proves is automatically absolute convergence. For a series with some negative terms, you can apply the integral test to the absolute values |aₙ| instead. If ∑|aₙ| converges by the integral test, then the original series converges absolutely.

Can the integral test be inconclusive?

When its conditions are met, the integral test always gives an answer, because the series and the integral share the same behavior. The test fails only when f is not positive, continuous and decreasing, or when the integral is too hard to evaluate. Keep in mind that it does not give the sum: ∫₁^∞ 1/x² dx = 1, but ∑ 1/n² = π²/6 ≈ 1.645.

Recommended EffortlessMath Books

For a precalc workbook that builds the function and limit skills the integral test relies on, the Pre-Calculus for Beginners covers exponential, logarithmic, and rational functions. For algebra foundation, see Algebra II for Beginners.

Related EffortlessMath Lessons

If a topic on this page feels rusty, these short lessons go deeper:

• Exponential and log integrals
• Properties of definite integrals
• Rules of integration
• Definite integrals and growth
• Trigonometric integrals