Question

The series $\sum_{n=1}^\infty (-1)^n \frac{1}{n^2}$ is}

• A. Conditionally convergent
• B. Absolutely convergent
• C. Oscillatory
• D. Divergent

Hint:

To distinguish between convergence types:
- \(\sum (-1)^n \frac{1}{n}\) is Conditionally Convergent (because \(\sum \frac{1}{n}\) diverges, but the alternating series converges).
- \(\sum (-1)^n \frac{1}{n^2}\) is Absolutely Convergent (because the absolute series \(\sum \frac{1}{n^2}\) converges).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
An infinite series \(\sum a_n\) is absolutely convergent if the series of its absolute values, \(\sum |a_n|\), is convergent.
If \(\sum |a_n|\) diverges but the alternating series \(\sum a_n\) converges, the series is conditionally convergent.

Step 2: Key Formula or Approach:
For the given alternating series:
\[ a_n = (-1)^n \frac{1}{n^2} \] We evaluate the absolute series:
\[ \sum_{n=1}^\infty |a_n| = \sum_{n=1}^\infty \frac{1}{n^2} \] We test this absolute series using the p-series test: the series \(\sum \frac{1}{n^p}\) converges if and only if \(p > 1\).

Step 3: Detailed Explanation:
Let us perform the convergence test step-by-step:
1. Extract the absolute values of the terms:
\[ |a_n| = \left| (-1)^n \frac{1}{n^2} \right| = \frac{1}{n^2} \] 2. Write the series of absolute values:
\[ \sum_{n=1}^\infty |a_n| = \sum_{n=1}^\infty \frac{1}{n^2} \] 3. Apply the p-series test to this series:
- Here, the exponent is \(p = 2\).
- Since \(2 > 1\), the series of absolute values \(\sum \frac{1}{n^2}\) is convergent.
4. Because the series of absolute values converges, the original alternating series \(\sum_{n=1}^\infty (-1)^n \frac{1}{n^2}\) is absolutely convergent by definition.
Therefore, the series is absolutely convergent.

Step 4: Final Answer:
The given infinite series is absolutely convergent, matching Option (B).