Question

The series \(\sum_{n=1}^\infty a_n = \sum_{n=1}^\infty \frac{1}{n^2}\) and \(\sum_{n=1}^\infty b_n = \sum_{n=1}^\infty \frac{1}{n^3}\) are

• A. Both divergent
• B. Both convergent
• C. \(\sum a_n\) is convergent and \(\sum b_n\) is divergent
• D. \(\sum a_n\) is divergent and \(\sum b_n\) is convergent

Hint:

Any series of the form \( \sum \frac{1}{n^p} \) where the exponent \( p \) is an integer greater than 1 (such as \( p = 2, 3, 4, \dots \)) is always convergent.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This question tests the convergence properties of multiple infinite series using the p-series test.

Step 2: Detailed Explanation:
Let us analyze each of the given infinite series:
- First Series: \( \sum_{n=1}^\infty a_n = \sum_{n=1}^\infty \frac{1}{n^2} \)
This is a p-series of the form \( \sum \frac{1}{n^p} \), where the exponent is \( p = 2 \).
According to the p-series test, the series converges if \( p > 1 \).
Since \( 2 > 1 \), the series \( \sum \frac{1}{n^2} \) is convergent.
- Second Series: \( \sum_{n=1}^\infty b_n = \sum_{n=1}^\infty \frac{1}{n^3} \)
This is also a p-series of the form \( \sum \frac{1}{n^p} \), where the exponent is \( p = 3 \).
Applying the same p-series test, the series converges if \( p > 1 \).
Since \( 3 > 1 \), the series \( \sum \frac{1}{n^3} \) is also convergent.
Therefore, both series are convergent.

Step 3: Final Answer:
Both series are convergent.
Therefore, the correct choice is Option (B).