Evaluate The Series Or State That It Diverges

Divergence and convergence are fundamental concepts in mathematics, especially within the study of infinite series. Determining whether a series converges or diverges is a crucial skill in calculus and real analysis. This article provides a comprehensive guide on how to evaluate series, detailing various tests and strategies to confidently assess their behavior.

Understanding Series Convergence and Divergence

An infinite series is the sum of an infinite number of terms. Represented as ∑ₙ₌₁^∞ aₙ, where aₙ represents the nth term of the series, determining whether this sum approaches a finite value (converges) or grows without bound (diverges) is essential.

• Convergence: A series converges if the sequence of its partial sums approaches a finite limit. Formally, if Sₙ = a₁ + a₂ + ... + aₙ, and lim (n→∞) Sₙ = L, where L is a finite number, the series converges to L.
• Divergence: A series diverges if the sequence of its partial sums does not approach a finite limit. This can occur in several ways: the partial sums may increase (or decrease) without bound, oscillate between two or more values, or behave erratically without settling down.

Evaluating series to determine convergence or divergence often involves employing various tests. Each test has specific conditions under which it is valid and useful.

Essential Tests for Evaluating Series

Numerous tests can be used to evaluate series, each suited to different types of series. Here are some of the most important:

1. The Divergence Test (nth-Term Test)

• Statement: If lim (n→∞) aₙ ≠ 0, then the series ∑ₙ₌₁^∞ aₙ diverges.
• Explanation: This test is straightforward and used as a preliminary check. If the terms of the series do not approach zero, the series cannot converge, as adding non-zero terms indefinitely will cause the sum to grow without bound.
• Example: Consider the series ∑ₙ₌₁^∞ (n / (n + 1)). Here, lim (n→∞) (n / (n + 1)) = 1, which is not zero. Therefore, by the Divergence Test, the series diverges.
• Caveat: The converse of the Divergence Test is not true. If lim (n→∞) aₙ = 0, it does not necessarily mean the series converges. Further testing is required.

2. The Integral Test

• Statement: If f(x) is a continuous, positive, and decreasing function on the interval [1, ∞), and aₙ = f(n) for all positive integers n, then the series ∑ₙ₌₁^∞ aₙ and the integral ∫₁^∞ f(x) dx either both converge or both diverge.
• Explanation: The Integral Test relates the convergence of a series to the convergence of an improper integral. If the integral converges, the series converges, and if the integral diverges, the series diverges.
• Example: Consider the series ∑ₙ₌₁^∞ (1 / n²). Let f(x) = 1 / x². The function f(x) is continuous, positive, and decreasing on [1, ∞). The integral ∫₁^∞ (1 / x²) dx = lim (b→∞) [-1/x]₁^b = lim (b→∞) (-1/b + 1) = 1, which is finite. Therefore, the series ∑ₙ₌₁^∞ (1 / n²) converges.
• Conditions: The function f(x) must meet all three criteria: continuity, positivity, and decreasing behavior for x ≥ 1.

3. The Comparison Test

• Statement: Suppose aₙ and bₙ are positive terms.
  • If aₙ ≤ bₙ for all n and ∑ₙ₌₁^∞ bₙ converges, then ∑ₙ₌₁^∞ aₙ also converges.
  • If aₙ ≥ bₙ for all n and ∑ₙ₌₁^∞ bₙ diverges, then ∑ₙ₌₁^∞ aₙ also diverges.
• Explanation: The Comparison Test compares a given series with a series whose convergence or divergence is known. If the terms of the given series are smaller than those of a convergent series, the given series also converges. Conversely, if the terms of the given series are larger than those of a divergent series, the given series also diverges.
• Example: Consider the series ∑ₙ₌₁^∞ (1 / (n³ + 1)). We can compare this series with the convergent series ∑ₙ₌₁^∞ (1 / n³). Since n³ + 1 > n³ for all n, we have 1 / (n³ + 1) < 1 / n³. Since ∑ₙ₌₁^∞ (1 / n³) converges (it’s a p-series with p = 3 > 1), by the Comparison Test, ∑ₙ₌₁^∞ (1 / (n³ + 1)) also converges.

4. The Limit Comparison Test

• Statement: Suppose aₙ > 0 and bₙ > 0 for all n. If lim (n→∞) (aₙ / bₙ) = c, where c is a finite number and c > 0, then both series ∑ₙ₌₁^∞ aₙ and ∑ₙ₌₁^∞ bₙ either both converge or both diverge.
• Explanation: The Limit Comparison Test is particularly useful when it is difficult to establish a direct inequality as required by the Comparison Test. It compares the limiting behavior of the ratio of the terms.
• Example: Consider the series ∑ₙ₌₁^∞ (n / (2n² + 1)). Let aₙ = n / (2n² + 1) and bₙ = 1 / n. Then, lim (n→∞) (aₙ / bₙ) = lim (n→∞) ((n / (2n² + 1)) / (1 / n)) = lim (n→∞) (n² / (2n² + 1)) = 1/2, which is a finite positive number. Since ∑ₙ₌₁^∞ (1 / n) is a divergent harmonic series, by the Limit Comparison Test, ∑ₙ₌₁^∞ (n / (2n² + 1)) also diverges.

5. The Ratio Test

• Statement: For the series ∑ₙ₌₁^∞ aₙ, let L = lim (n→∞) |aₙ₊₁ / aₙ|.
  • If L < 1, the series converges absolutely.
  • If L > 1 (or L = ∞), the series diverges.
  • If L = 1, the test is inconclusive.
• Explanation: The Ratio Test is particularly useful for series involving factorials or exponential terms. It examines the ratio of consecutive terms to determine convergence.
• Example: Consider the series ∑ₙ₌₁^∞ (n! / nⁿ). Let aₙ = n! / nⁿ. Then aₙ₊₁ = (n+1)! / (n+1)ⁿ⁺¹. So, L = lim (n→∞) |aₙ₊₁ / aₙ| = lim (n→∞) |((n+1)! / (n+1)ⁿ⁺¹) / (n! / nⁿ)| = lim (n→∞) |((n+1) / (n+1)ⁿ⁺¹) * nⁿ| = lim (n→∞) |nⁿ / (n+1)ⁿ| = lim (n→∞) |1 / (1 + 1/n)ⁿ| = 1/e. Since 1/e < 1, the series converges absolutely.

6. The Root Test

• Statement: For the series ∑ₙ₌₁^∞ aₙ, let L = lim (n→∞) |aₙ|^(1/n).
  • If L < 1, the series converges absolutely.
  • If L > 1 (or L = ∞), the series diverges.
  • If L = 1, the test is inconclusive.
• Explanation: The Root Test is particularly useful for series where the nth term involves an nth power.
• Example: Consider the series ∑ₙ₌₁^∞ ( (2n + 3) / (3n + 2) )ⁿ. Let aₙ = ((2n + 3) / (3n + 2))ⁿ. Then, L = lim (n→∞) |aₙ|^(1/n) = lim (n→∞) |((2n + 3) / (3n + 2))ⁿ|^(1/n) = lim (n→∞) |(2n + 3) / (3n + 2)| = 2/3. Since 2/3 < 1, the series converges absolutely.

7. Alternating Series Test

• Statement: If the alternating series ∑ₙ₌₁^∞ (-1)ⁿ⁻¹ bₙ satisfies the following conditions:
  • bₙ > 0 for all n,
  • bₙ is a decreasing sequence, i.e., bₙ₊₁ ≤ bₙ for all n,
  • lim (n→∞) bₙ = 0, then the series converges.
• Explanation: The Alternating Series Test applies to series with alternating signs. It provides conditions under which such a series converges.
• Example: Consider the alternating series ∑ₙ₌₁^∞ (-1)ⁿ⁻¹ / n. Here, bₙ = 1 / n.
  • bₙ > 0 for all n.
  • bₙ₊₁ = 1 / (n+1) ≤ 1 / n = bₙ, so the sequence is decreasing.
  • lim (n→∞) bₙ = lim (n→∞) (1 / n) = 0. Since all three conditions are satisfied, the series converges.

8. Absolute and Conditional Convergence

• Absolute Convergence: A series ∑ₙ₌₁^∞ aₙ converges absolutely if the series ∑ₙ₌₁^∞ |aₙ| converges.
• Conditional Convergence: A series ∑ₙ₌₁^∞ aₙ converges conditionally if it converges but does not converge absolutely.
• Explanation: Absolute convergence implies that the series converges regardless of the signs of its terms. Conditional convergence, on the other hand, means that the convergence depends on the alternating signs.
• Example: The series ∑ₙ₌₁^∞ (-1)ⁿ⁻¹ / n converges conditionally because it converges, but ∑ₙ₌₁^∞ |(-1)ⁿ⁻¹ / n| = ∑ₙ₌₁^∞ (1 / n) diverges (harmonic series).

Strategies for Evaluating Series

Evaluating series requires a strategic approach. Here’s a step-by-step guide:

1. Preliminary Check:
   • Divergence Test: Always start with the Divergence Test. If lim (n→∞) aₙ ≠ 0, the series diverges, and no further testing is needed.
2. Identify the Series Type:
   • Geometric Series: If the series is of the form ∑ₙ₌₀^∞ arⁿ, where a is a constant and r is the common ratio, it converges if |r| < 1 and diverges if |r| ≥ 1.
   • p-Series: If the series is of the form ∑ₙ₌₁^∞ 1 / nᵖ, it converges if p > 1 and diverges if p ≤ 1.
   • Alternating Series: If the series has alternating signs, consider using the Alternating Series Test.
   • Series with Factorials or Exponentials: If the series involves factorials or exponential terms, the Ratio Test is often effective.
   • Series with nth Powers: If the series involves nth powers, the Root Test may be useful.
3. Choose an Appropriate Test:
   • Based on the identified series type and the form of the terms, select the most appropriate test.
   • Consider the conditions required for each test to be valid.
4. Apply the Test:
   • Carefully apply the selected test, showing all steps in the process.
   • Pay attention to limits and inequalities.
5. Draw a Conclusion:
   • Based on the results of the test, conclude whether the series converges or diverges.
   • If the test is inconclusive, try a different test.

Examples of Series Evaluation

Let's walk through several examples to illustrate the application of these tests:

Example 1: Evaluate the series ∑ₙ₌₁^∞ (n² / 2ⁿ).

• Preliminary Check: lim (n→∞) (n² / 2ⁿ) = 0, so the Divergence Test is inconclusive.
• Series Type: The series involves an exponential term (2ⁿ), so the Ratio Test may be effective.
• Ratio Test:
  • Let aₙ = n² / 2ⁿ.
  • aₙ₊₁ = (n+1)² / 2ⁿ⁺¹.
  • L = lim (n→∞) |aₙ₊₁ / aₙ| = lim (n→∞) |((n+1)² / 2ⁿ⁺¹) / (n² / 2ⁿ)| = lim (n→∞) |((n+1)² / n²) * (2ⁿ / 2ⁿ⁺¹)| = lim (n→∞) |((n+1)² / n²) * (1/2)| = lim (n→∞) |(n² + 2n + 1) / (2n²)| = 1/2.
• Conclusion: Since L = 1/2 < 1, the series converges absolutely.

Example 2: Evaluate the series ∑ₙ₌₁^∞ (1 / (n√(n))).

• Preliminary Check: lim (n→∞) (1 / (n√(n))) = 0, so the Divergence Test is inconclusive.
• Series Type: This is a p-series with p = 3/2.
• p-Series Test: The series can be written as ∑ₙ₌₁^∞ (1 / n^(3/2)). Since p = 3/2 > 1, the series converges.
• Conclusion: The series converges.

Example 3: Evaluate the series ∑ₙ₌₁^∞ (-1)ⁿ / (2n + 1).

• Preliminary Check: lim (n→∞) (-1)ⁿ / (2n + 1) = 0, so the Divergence Test is inconclusive.
• Series Type: This is an alternating series.
• Alternating Series Test:
  • bₙ = 1 / (2n + 1) > 0 for all n.
  • bₙ₊₁ = 1 / (2(n+1) + 1) = 1 / (2n + 3) ≤ 1 / (2n + 1) = bₙ, so the sequence is decreasing.
  • lim (n→∞) bₙ = lim (n→∞) (1 / (2n + 1)) = 0.
• Conclusion: Since all three conditions are satisfied, the series converges.

Example 4: Evaluate the series ∑ₙ₌₁^∞ (3ⁿ / n!).

• Preliminary Check: lim (n→∞) (3ⁿ / n!) = 0, so the Divergence Test is inconclusive.
• Series Type: The series involves a factorial, so the Ratio Test may be effective.
• Ratio Test:
  • Let aₙ = 3ⁿ / n!.
  • aₙ₊₁ = 3ⁿ⁺¹ / (n+1)!.
  • L = lim (n→∞) |aₙ₊₁ / aₙ| = lim (n→∞) |(3ⁿ⁺¹ / (n+1)!) / (3ⁿ / n!)| = lim (n→∞) |(3ⁿ⁺¹ / 3ⁿ) * (n! / (n+1)!)| = lim (n→∞) |3 / (n+1)| = 0.
• Conclusion: Since L = 0 < 1, the series converges absolutely.

Common Mistakes to Avoid

When evaluating series, it’s essential to avoid common mistakes:

• Incorrect Application of Tests: Ensure that the conditions for each test are met before applying it. For example, the Integral Test requires a continuous, positive, and decreasing function.
• Misinterpreting the Divergence Test: Remember that if lim (n→∞) aₙ = 0, it does not imply convergence. Further testing is required.
• Algebraic Errors: Be meticulous with algebraic manipulations when calculating limits and ratios.
• Forgetting to Check Conditions for Alternating Series Test: Ensure that bₙ > 0, bₙ is decreasing, and lim (n→∞) bₙ = 0.
• Not Identifying the Series Type: Failing to recognize the type of series (e.g., geometric, p-series, alternating) can lead to choosing an inappropriate test.

Conclusion

Evaluating series to determine convergence or divergence is a fundamental skill in calculus and real analysis. Mastering the various tests and strategies outlined in this article will enable you to confidently assess the behavior of infinite series. Remember to start with the Divergence Test, identify the series type, choose an appropriate test, apply it carefully, and draw a logical conclusion. By understanding and applying these techniques, you can successfully navigate the complexities of series convergence and divergence.