Math Problem Statement

does ((-1)^n)/(5n+1) converge or diverge

Solution

To determine whether the series n=0(1)n5n+1\sum_{n=0}^{\infty} \frac{(-1)^n}{5n + 1} converges or diverges, we can use the Alternating Series Test.

  1. Identify the terms: The terms of the series are an=15n+1a_n = \frac{1}{5n + 1}.

  2. Check the conditions for the Alternating Series Test:

    • Condition 1: ana_n must be positive. Since 5n+1>05n + 1 > 0 for all n0n \geq 0, ana_n is positive.

    • Condition 2: ana_n must be decreasing. We can show that ana_n is decreasing by examining anan+1a_n - a_{n+1}: anan+1=15n+115(n+1)+1=15n+115n+6a_n - a_{n+1} = \frac{1}{5n + 1} - \frac{1}{5(n + 1) + 1} = \frac{1}{5n + 1} - \frac{1}{5n + 6} The common denominator is (5n+1)(5n+6)(5n + 1)(5n + 6), leading to: anan+1=(5n+6)(5n+1)(5n+1)(5n+6)=5(5n+1)(5n+6)>0a_n - a_{n+1} = \frac{(5n + 6) - (5n + 1)}{(5n + 1)(5n + 6)} = \frac{5}{(5n + 1)(5n + 6)} > 0 Hence, ana_n is decreasing.

    • Condition 3: The limit of ana_n as nn approaches infinity must be zero: limnan=limn15n+1=0\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{1}{5n + 1} = 0

Since all three conditions of the Alternating Series Test are satisfied, the series converges.

Conclusion

The series n=0(1)n5n+1\sum_{n=0}^{\infty} \frac{(-1)^n}{5n + 1} converges.

Would you like further details or have any questions about this process? Here are some related questions to consider:

  1. What is the difference between absolute and conditional convergence?
  2. How would you determine the sum of this series if it converges?
  3. Can you apply other convergence tests to this series?
  4. What happens if the series had a different alternating factor?
  5. How does the convergence of this series compare to the convergence of a related series?

Tip: Always check the conditions of the test you are using carefully to ensure proper conclusions about convergence.

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Math Problem Analysis

Mathematical Concepts

Series Convergence
Alternating Series

Formulas

Alternating Series Test

Theorems

Alternating Series Test

Suitable Grade Level

Grades 11-12