Comparison Test And Limit Comparison Test

Comparison Test and Limit Comparison Test: A Comprehensive Guide

Determining the convergence or divergence of an infinite series is a fundamental problem in calculus. While some series yield to simple tests, many require more sophisticated techniques. Two powerful tools in this arsenal are the Comparison Test and the Limit Comparison Test. This comprehensive guide will delve into both tests, exploring their applications, limitations, and providing numerous examples to solidify your understanding.

Understanding the Comparison Test

The Comparison Test is a direct method for determining the convergence or divergence of a series by comparing it to a series whose convergence or divergence is already known. The intuition is straightforward: if a series is term-by-term smaller than a known convergent series, it must also converge. Conversely, if a series is term-by-term larger than a known divergent series, it must also diverge.

Formal Statement of the Comparison Test:

Let Σa_n and Σb_n be two series with non-negative terms (a_n ≥ 0 and b_n ≥ 0 for all n).

• Case 1 (Direct Comparison): If 0 ≤ a_n ≤ b_n for all n, then:

  • If Σb_n converges, then Σa_n converges.
  • If Σa_n diverges, then Σb_n diverges.

• Case 2 (Reverse Comparison - less common but equally valid): If 0 ≤ b_n ≤ a_n for all n, then:

  • If Σb_n diverges, then Σa_n diverges.
  • If Σa_n converges, then Σb_n converges.

Key Considerations for the Comparison Test:

• Non-negativity: The Comparison Test only applies to series with non-negative terms. For series with mixed signs, other tests are necessary (like the alternating series test).
• Finding a suitable comparison series: The success of the Comparison Test hinges on finding a suitable series (Σb_n) whose convergence or divergence is known. This often requires experience and familiarity with common convergent and divergent series (e.g., geometric series, p-series).
• Inequality: Establishing the inequality (a_n ≤ b_n or b_n ≤ a_n) is crucial. This often involves algebraic manipulation and careful consideration of the terms.

Example 1: Using the Comparison Test to prove convergence

Consider the series Σ(1/(n² + 1)). We suspect this converges. Let's compare it to the convergent p-series Σ(1/n²) (p=2 > 1).

For all n ≥ 1, we have 0 ≤ 1/(n² + 1) ≤ 1/n². Since Σ(1/n²) converges, by the Comparison Test, Σ(1/(n² + 1)) also converges.

Example 2: Using the Comparison Test to prove divergence

Consider the series Σ(1/(n + 1)). Let's compare it to the harmonic series Σ(1/n), which is known to diverge.

For all n ≥ 1, we have 0 ≤ 1/n ≤ 1/(n + 1). However, this inequality is reversed compared to the typical Comparison Test setup. But, since Σ(1/n) diverges, the reverse comparison tells us that Σ(1/(n+1)) also diverges.

The Limit Comparison Test: A More Refined Approach

The Limit Comparison Test offers a more flexible alternative to the Comparison Test. It's particularly useful when the inequality required for the direct comparison is difficult to establish directly.

Formal Statement of the Limit Comparison Test:

Let Σa_n and Σb_n be two series with positive terms (a_n > 0 and b_n > 0 for all n). If the limit

lim (n→∞) (a_n/b_n) = L

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

Key Advantages of the Limit Comparison Test:

• Flexibility: It doesn't require a strict inequality between the terms, only that their ratio approaches a finite positive limit. This makes it easier to find a suitable comparison series.
• Efficiency: It often simplifies the comparison process, especially when dealing with complex expressions.

Example 3: Using the Limit Comparison Test

Consider the series Σ( (3n² + 2n + 1) / (n⁵ + 4n³ + 7) ). Finding a suitable series for direct comparison might be challenging. Let's use the Limit Comparison Test.

We can compare it to Σ(1/n³), a convergent p-series. Let's compute the limit:

lim (n→∞) [ (3n² + 2n + 1) / (n⁵ + 4n³ + 7) ] / (1/n³) = lim (n→∞) (3n⁵ + 2n⁴ + n³) / (n⁵ + 4n³ + 7) = 3

Since the limit is 3 (a finite positive number), and Σ(1/n³) converges, by the Limit Comparison Test, Σ( (3n² + 2n + 1) / (n⁵ + 4n³ + 7) ) also converges.

Example 4: Using the Limit Comparison Test for Divergence

Let's examine the series Σ( (√(n² + 1)) / (n + 2) ). We can compare it to the divergent series Σ(1/n) (harmonic series). Let's evaluate the limit:

lim (n→∞) [ (√(n² + 1)) / (n + 2) ] / (1/n) = lim (n→∞) [ n√(n² + 1) ] / (n + 2) = lim (n→∞) [ n²/n] = ∞.

The limit is ∞, not a finite positive number, and so the Limit Comparison Test doesn't provide conclusive information here. It may still diverge, however, this example highlights the fact that only a finite positive limit provides decisive information. We could choose a different comparison to prove its divergence via the LCT. Let's try comparing to Σ(1/√n), which is a divergent p-series (p = 1/2 ≤ 1).

lim (n→∞) [ (√(n² + 1)) / (n + 2) ] / (1/√n) = lim (n→∞) √(n(n² + 1)) / (n + 2) = ∞

Again, we don't get a finite positive limit. In this case, direct comparison will probably be more useful.

Choosing Between the Comparison Test and the Limit Comparison Test

Both tests are valuable tools, but choosing the right one depends on the specific series.

• Use the Comparison Test when:

  • A simple, direct inequality is readily apparent between the series terms and a known convergent/divergent series.
  • The series terms have relatively simple expressions allowing direct comparisons

• Use the Limit Comparison Test when:

  • Establishing a direct inequality is difficult or cumbersome.
  • The series terms are complex and the Limit Comparison Test simplifies the comparison process significantly.
  • you are more comfortable with finding limits than directly manipulating inequalities

Conclusion

The Comparison Test and the Limit Comparison Test are fundamental tools for determining the convergence or divergence of infinite series. Mastering these tests empowers you to tackle a wide range of problems in calculus and beyond. While the Comparison Test offers a direct and intuitive approach, the Limit Comparison Test provides a more flexible and often more efficient method for analyzing complex series. By understanding the strengths and limitations of each test, you can confidently select the most appropriate method for a given problem, and significantly expand your abilities to solve problems involving series. Remember to always carefully consider the conditions for each test and practice applying them to various examples to develop your intuition and proficiency.