Alternating Series Test Practice Problems

Mastering the Alternating Series Test: Practice Problems and Solutions

The Alternating Series Test is a crucial tool in the calculus arsenal for determining the convergence of infinite series. Understanding this test, however, requires more than just memorizing the criteria; it demands practice and a deep understanding of its underlying principles. This article provides a comprehensive guide, equipping you with the knowledge and practice problems to master the Alternating Series Test. We'll delve into the theorem itself, explore various examples with detailed solutions, and address common pitfalls. By the end, you’ll be confident in applying this test to a wide range of alternating series.

Understanding the Alternating Series Test Theorem

Before diving into problems, let's formally state the Alternating Series Test (AST):

Theorem: An alternating series of the form ∑ (-1)^n * b_n (where b_n ≥ 0 for all n) converges if:

1. b_(n+1) ≤ b_n for all n ≥ N, for some integer N. This means the terms are eventually monotonically decreasing (or at least non-increasing). Note that it doesn't require strict decrease for all n, just eventually.

2. lim (n→∞) b_n = 0. This means the terms approach zero as n goes to infinity.

If either of these conditions is not met, the test is inconclusive. The series may still converge, but the AST cannot confirm it. It's important to remember that the AST only provides a sufficient condition for convergence, not a necessary one.

Practice Problems: From Easy to Advanced

Let's tackle a range of problems, starting with simpler examples and progressively increasing the complexity. Each problem will be solved step-by-step, highlighting the application of the Alternating Series Test and crucial considerations.

Problem 1 (Easy): Determine the convergence of the series ∑ (-1)^n (1/n).

Solution:

This is the well-known alternating harmonic series.

1. Monotonicity: b_n = 1/n. Clearly, 1/(n+1) ≤ 1/n for all n ≥ 1. The terms are monotonically decreasing.

2. Limit of b_n: lim (n→∞) (1/n) = 0.

Since both conditions are met, the Alternating Series Test confirms that the series converges.

Problem 2 (Medium): Investigate the convergence of ∑ (-1)^n (n/(n^2 + 1)).

Solution:

1. Monotonicity: Let's analyze the function f(x) = x/(x² + 1). Taking the derivative, we get f'(x) = (1 - x²)/(x² + 1)². For x > 1, f'(x) < 0, indicating that the function is decreasing for x > 1. Thus, b_(n+1) ≤ b_n for sufficiently large n.

2. Limit of b_n: lim (n→∞) (n/(n² + 1)) = lim (n→∞) (1/(n + 1/n)) = 0.

Both conditions are satisfied, so the series converges by the Alternating Series Test.

Problem 3 (Medium): Determine whether the series ∑ (-1)^n (n! / n^n) converges.

Solution:

1. Monotonicity: Let's consider the ratio of consecutive terms:

   b_(n+1) / b_n = [(n+1)! / (n+1)^(n+1)] / [n! / n^n] = [(n+1)n! / (n+1)^(n+1)] * [n^n / n!] = n^n / (n+1)^n = (n/(n+1))^n = (1/(1 + 1/n))^n

   As n approaches infinity, this ratio approaches 1/e < 1. This implies that the terms are eventually decreasing.

2. Limit of b_n: Using the ratio test (or recognizing the limit as related to 1/e), we find that lim (n→∞) (n! / n^n) = 0.

Therefore, the series converges by the Alternating Series Test.

Problem 4 (Hard): Analyze the convergence of ∑ (-1)^n (ln n / n).

Solution:

1. Monotonicity: Let f(x) = ln x / x. Then f'(x) = (1 - ln x) / x². For x > e, f'(x) < 0, meaning the function is decreasing for x > e. Hence, the terms are eventually decreasing.

2. Limit of b_n: Using L'Hôpital's Rule, we find lim (n→∞) (ln n / n) = lim (n→∞) (1/n) = 0.

Both conditions are met; thus, the series converges by the Alternating Series Test.

Problem 5 (Hard): Investigate the convergence of ∑ (-1)^n (√n / (n + 1)).

Solution:

1. Monotonicity: Consider the function f(x) = √x / (x + 1). Taking the derivative, we get a somewhat complex expression. Instead, let's examine the ratio of consecutive terms:

   b_(n+1) / b_n = [√(n+1) / (n+2)] / [√n / (n+1)] = √[(n+1)/n] * [(n+1)/(n+2)]

   As n approaches infinity, this ratio approaches 1. This is inconclusive for monotonicity. However, we can analyze the function directly. We have f(x) = √x / (x+1). We look for x where f'(x) < 0. Let's simplify the derivative by using Wolfram Alpha or a similar tool. The derivative is always negative for sufficiently large x. Therefore the terms eventually decrease.

2. Limit of b_n: lim (n→∞) (√n / (n + 1)) = lim (n→∞) (1/√n * 1/(1+1/n)) = 0.

Therefore, the series converges according to the Alternating Series Test.

Problem 6 (Advanced): Determine the convergence of ∑ (-1)^n * (sin(1/n)).

Solution:

1. Monotonicity: The function f(x) = sin(1/x) is a decreasing function for x > 0. To see this consider its derivative:

f'(x) = cos(1/x) * (-1/x^2) which is negative for all x > 0 since cos(1/x) is always between -1 and 1, while -1/x^2 is negative. Hence, the terms are monotonically decreasing.

2. Limit of b_n: lim (n→∞) sin(1/n) = sin(0) = 0.

Since both conditions are fulfilled, the series converges using the Alternating Series Test.

Common Mistakes and Pitfalls

• Forgetting the Monotonicity Condition: Many students focus solely on the limit condition, overlooking the crucial requirement of eventually monotonically decreasing terms.

• Incorrectly Assessing Monotonicity: Determining monotonicity can be challenging. Always be rigorous in your analysis, using derivatives or ratio tests when necessary.

• Assuming the Test is Necessary and Sufficient: The AST is only a sufficient condition. If the test fails, the series might still converge (though a different test would be needed).

• Misinterpreting Inconclusive Results: If either condition isn't met, the AST provides no information about the series' convergence.

Frequently Asked Questions (FAQ)

Q: What if the limit of b_n is not 0?

A: If lim (n→∞) b_n ≠ 0, the alternating series diverges. The terms don't approach zero, preventing convergence.

Q: Can I use the Alternating Series Test for non-alternating series?

A: No. The AST is specifically designed for series with alternating signs.

Q: What if the terms are not monotonically decreasing from the beginning?

A: The condition only requires the terms to be eventually monotonically decreasing (non-increasing). There might be a finite number of terms that don’t follow the pattern.

Q: What if the series is not in the standard alternating form?

A: You might be able to rearrange the series to fit the required form or use a different convergence test.

Q: How do I determine the remainder in an alternating series?

A: The remainder of an alternating series is bounded by the next term in the sequence (the absolute value of the next term). This is a useful property of convergent alternating series.

Conclusion

The Alternating Series Test is a powerful tool for determining the convergence of alternating series. Mastering this test requires understanding its conditions, practicing with diverse problems, and recognizing potential pitfalls. By diligently working through the examples and understanding the underlying principles, you’ll develop the confidence to tackle even the most complex alternating series problems. Remember to always carefully check both the monotonicity and limit conditions. With practice, you will master this important concept in calculus and strengthen your analytical skills.