Comparison Test For Infinite Series

A Comprehensive Guide to Comparison Tests for Infinite Series

Determining the convergence or divergence of an infinite series is a fundamental concept in calculus. While some series yield easily to direct application of tests like the geometric series test or the p-series test, many others require more sophisticated approaches. Among these, comparison tests prove invaluable for analyzing the convergence behavior of a wide range of series. This article provides a thorough exploration of different comparison tests, including the direct comparison test and the limit comparison test, offering practical examples and explanations to solidify understanding.

Introduction: Understanding Infinite Series and Convergence

An infinite series is the sum of an infinite sequence of numbers. We represent it as:

∑_n=1^∞ a_n = a₁ + a₂ + a₃ + ...

The crucial question is whether this sum converges to a finite value or diverges to infinity (or oscillates). If the sequence of partial sums (S_n = a₁ + a₂ + ... + a_n) approaches a limit L as n approaches infinity, the series converges to L. Otherwise, it diverges. Comparison tests offer a powerful way to determine convergence without needing to find the exact sum.

The Direct Comparison Test: A Simple Yet Powerful Tool

The direct comparison test is based on the intuitive idea that if the terms of a series are consistently smaller than the terms of a known convergent series, then the original series must also converge. Conversely, if the terms are consistently larger than the terms of a known divergent series, the original series must also diverge.

Theorem (Direct Comparison Test):

Let ∑_n=1^∞ a_n and ∑_n=1^∞ b_n be two series with positive terms (a_n ≥ 0 and b_n ≥ 0 for all n).

1. If ∑_n=1^∞ b_n converges and 0 ≤ a_n ≤ b_n for all n ≥ N (for some integer N), then ∑_n=1^∞ a_n converges.

2. If ∑_n=1^∞ b_n diverges and a_n ≥ b_n ≥ 0 for all n ≥ N, then ∑_n=1^∞ a_n diverges.

Example 1: Convergent Series

Consider the series ∑_n=1^∞ 1/(n² + 1). We can compare it to the convergent p-series ∑_n=1^∞ 1/n² (p = 2 > 1). Since 1/(n² + 1) < 1/n² for all n ≥ 1, by the direct comparison test, ∑_n=1^∞ 1/(n² + 1) converges.

Example 2: Divergent Series

Consider the series ∑_n=1^∞ (n + 1)/n. We can compare it to the divergent harmonic series ∑_n=1^∞ 1/n. Since (n + 1)/n = 1 + 1/n > 1/n for all n ≥ 1, the series ∑_n=1^∞ (n + 1)/n diverges by the direct comparison test.

Limitations of the Direct Comparison Test:

The direct comparison test is straightforward, but it's not always easy to find a suitable comparison series. Sometimes, the inequality required by the test might be difficult or impossible to establish directly. This is where the limit comparison test comes in handy.

The Limit Comparison Test: Refining the Comparison Approach

The limit comparison test addresses the limitations of the direct comparison test by focusing on the asymptotic behavior of the terms of the series. Instead of requiring a strict inequality between the terms, it examines the limit of the ratio of the terms.

Theorem (Limit Comparison Test):

Let ∑_n=1^∞ a_n and ∑_n=1^∞ b_n be two series with positive terms (a_n > 0 and b_n > 0 for all n). If the limit:

L = lim_n→∞ (a_n/b_n)

exists and is a finite positive number (0 < L < ∞), then both series either converge together or diverge together.

Example 3: Using the Limit Comparison Test

Let's consider the series ∑_n=1^∞ (3n² + 2n)/(n⁴ + 5). Finding a direct comparison might be challenging. However, we can use the limit comparison test. We observe that the dominant terms in the numerator and denominator are 3n² and n⁴, respectively. Let's compare it to the series ∑_n=1^∞ 3n²/n⁴ = ∑_n=1^∞ 3/n². This is a convergent p-series (p = 2).

Now, let's find the limit:

L = lim_n→∞ [(3n² + 2n)/(n⁴ + 5)] / (3/n²) = lim_n→∞ (3n⁴ + 2n³)/(3n⁴ + 15) = 1

Since L = 1 (a finite positive number), and ∑_n=1^∞ 3/n² converges, then by the limit comparison test, ∑_n=1^∞ (3n² + 2n)/(n⁴ + 5) also converges.

Example 4: Divergent Series using Limit Comparison Test

Consider the series ∑_n=1^∞ (√(n+1))/n. We can compare it to the harmonic series, ∑_n=1^∞ 1/√n which is a divergent p-series (p = 1/2 ≤ 1). Let's compute the limit:

L = lim_n→∞ [√(n+1)/n] / (1/√n) = lim_n→∞ √(n(n+1))/n = lim_n→∞ √(1 + 1/n) = 1

Since L = 1 and ∑_n=1^∞ 1/√n diverges, the series ∑_n=1^∞ (√(n+1))/n also diverges.

Comparison Tests and Alternating Series

The direct and limit comparison tests are primarily designed for series with positive terms. However, for alternating series (series where terms alternate in sign), other tests like the alternating series test are more appropriate. Comparison tests can sometimes be used indirectly to determine the absolute convergence of an alternating series. If the absolute values of the terms of an alternating series form a convergent series (using a comparison test), then the alternating series is absolutely convergent, and therefore convergent.

Frequently Asked Questions (FAQ)

Q1: What if the limit in the limit comparison test is 0 or ∞?

A1: If the limit L is 0, the test is inconclusive. If L is ∞, and the comparison series diverges, then the original series diverges. However, if L is ∞ and the comparison series converges, the test is inconclusive.

Q2: Can I use the comparison tests for series with negative terms?

A2: The direct and limit comparison tests are primarily for series with positive terms. If you have a series with some negative terms, consider first checking for absolute convergence. If the absolute value of the terms converges, the original series also converges. If the absolute value of the terms diverges, you might need to consider other convergence tests.

Q3: Which comparison test should I use?

A3: The direct comparison test is simpler but requires finding a comparison series that clearly satisfies the inequality condition. The limit comparison test is more flexible and often easier to apply, but it requires calculating a limit. Choose the test that seems more easily applicable based on the series you are analyzing.

Conclusion: Mastering Comparison Tests for Series Analysis

Comparison tests, both direct and limit, are powerful tools in the arsenal of techniques for determining the convergence or divergence of infinite series. Understanding their application and limitations is crucial for effectively analyzing a wide range of series. By carefully selecting the appropriate comparison series and applying the theorems correctly, you can confidently determine the convergence behavior of many challenging series. Remember to always check the conditions of the tests before applying them, ensuring the positivity of terms (for direct and limit comparison tests) and considering alternative tests for series with negative terms or alternating series. Mastering these techniques allows for a deeper understanding of infinite series and their role in various branches of mathematics and science.