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

1. **Identify the terms:** The terms of the series are \(a_n = \frac{1}{5n + 1}\).

2. **Check the conditions for the Alternating Series Test:**
   - **Condition 1:** \(a_n\) must be positive. Since \(5n + 1 > 0\) for all \(n \geq 0\), \(a_n\) is positive.
   - **Condition 2:** \(a_n\) must be decreasing. We can show that \(a_n\) is decreasing by examining \(a_n - a_{n+1}\):
     \[
     a_n - a_{n+1} = \frac{1}{5n + 1} - \frac{1}{5(n + 1) + 1} = \frac{1}{5n + 1} - \frac{1}{5n + 6}
     \]
     The common denominator is \((5n + 1)(5n + 6)\), leading to:
     \[
     a_n - a_{n+1} = \frac{(5n + 6) - (5n + 1)}{(5n + 1)(5n + 6)} = \frac{5}{(5n + 1)(5n + 6)} > 0
     \]
     Hence, \(a_n\) is decreasing.

   - **Condition 3:** The limit of \(a_n\) as \(n\) approaches infinity must be zero:
     \[
     \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{1}{5n + 1} = 0
     \]

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

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

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

Explore the convergence of the series \(\sum_{n=0}^{\infty} \frac{(-1)^n}{5n + 1}\) using the Alternating Series Test. This problem demonstrates key mathematical concepts and provides a clear step-by-step solution. Suitable for students in Grades 11-12.

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