Question

Let $\sum_{n=1}^\infty a_n$ is absolutely convergent series, then}

• A. $\sum_{n=1}^\infty a_n$ is convergent
• B. $\sum_{n=1}^\infty a_n$ is conditionally convergent
• C. $\sum_{n=1}^\infty a_n$ is not convergent
• D. $\sum_{n=1}^\infty a_n$ is oscillatory

Hint:

Remember the direction of this logical implication:
- Absolute Convergence \(\implies\) Convergence (always true).
- The reverse is not always true: a series can converge without converging absolutely (which defines Conditional Convergence).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
In the study of infinite series, we analyze different types of convergence.
The Absolute Convergence Theorem defines the relationship between absolute convergence and standard convergence.

Step 2: Key Formula or Approach:
The Absolute Convergence Theorem states:
\[ \text{If } \sum_{n=1}^\infty |a_n| \text{ converges, then } \sum_{n=1}^\infty a_n \text{ must also converge.} \]

Step 3: Detailed Explanation:
Let us prove this theorem using the triangle inequality:
1. We are given that \(\sum |a_n|\) converges.
2. For any real numbers, the inequality holds:
\[ 0 \le a_n + |a_n| \le 2|a_n| \] 3. Because \(\sum |a_n|\) converges, the series \(\sum 2|a_n|\) also converges.
4. Using the Direct Comparison Test, the series \(\sum (a_n + |a_n|)\) must also converge because its terms are bounded by the convergent series \(\sum 2|a_n|\).
5. We can write the term \(a_n\) as:
\[ a_n = (a_n + |a_n|) - |a_n| \] 6. Because the difference of two convergent series is always convergent:
\[ \sum_{n=1}^\infty a_n = \sum_{n=1}^\infty (a_n + |a_n|) - \sum_{n=1}^\infty |a_n| \] This proves that the series \(\sum a_n\) is convergent.
Therefore, any absolutely convergent series is always convergent.
- Let us evaluate the incorrect options:
- Option (B): A series cannot be both conditionally convergent and absolutely convergent; these are mutually exclusive classifications.
- Option (C) and (D) contradict this theorem.

Step 4: Final Answer:
Every absolutely convergent series is convergent, matching Option (A).