Question

Let \(\sum_{n=1}^{\infty}(-1)^na_n=\sum_{n=1}^{\infty}(-1)^n\dfrac{1}{\sqrt{n}}\) be a series, then:

• A. $\sum_{\text{n}=1}^{\infty} (-1)^\text{n} \text{a}_\text{n}$ is divergent
• B. $\sum_{\text{n}=1}^{\infty} (\text{a}_\text{n})^2$ is absolutely convergent
• C. $\sum_{\text{n}=1}^{\infty} (\text{a}_\text{n})^3$ is convergent
• D. $\sum_{\text{n}=1}^{\infty} \text{a}_\text{n}$ is convergent series

Hint:

Always use the p-series test to check convergence of series containing terms like \( n^k \):
- \( \sum \frac{1}{n} \) (harmonic series) is divergent.
- \( \sum \frac{1}{n^{1.5}} \) is convergent.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We evaluate the convergence of different series derived from the term \( a_n = \frac{1}{\sqrt{n}} \) using standard series tests, such as the p-series test and the Alternating Series Test (Leibniz's Test).
Key Formula or Approach:
The p-series test states that the series \( \sum_{n=1}^{\infty} \frac{1}{n^p} \) is:
- Convergent if \( p > 1 \)
- Divergent if \( p \le 1 \)

Step 2: Detailed Explanation:
Given the alternating series \( \sum_{n=1}^{\infty} (-1)^n a_n \) with:
\[ a_n = \frac{1}{\sqrt{n}} = \frac{1}{n^{1/2}} \]
Let us analyze each option:
- Analyzing Option A:
By Leibniz's Test for alternating series, since \( a_n > 0 \), \( a_n \) is strictly decreasing, and \( \lim_{n\to\infty} a_n = 0 \), the alternating series \( \sum_{n=1}^{\infty} (-1)^n a_n \) is convergent.
Thus, Option A is false.
- Analyzing Option B:
We find \( (a_n)^2 \):
\[ (a_n)^2 = \left( \frac{1}{\sqrt{n}} \right)^2 = \frac{1}{n} \]
The series is \( \sum_{n=1}^{\infty} (a_n)^2 = \sum_{n=1}^{\infty} \frac{1}{n} \), which is the classic harmonic series.
Since \( p = 1 \), this series is divergent by the p-series test, which means it cannot be absolutely convergent.
Thus, Option B is false.
- Analyzing Option C:
We find \( (a_n)^3 \):
\[ (a_n)^3 = \left( \frac{1}{\sqrt{n}} \right)^3 = \frac{1}{n^{3/2}} \]
The series is \( \sum_{n=1}^{\infty} (a_n)^3 = \sum_{n=1}^{\infty} \frac{1}{n^{1.5}} \).
By the p-series test, since \( p = 1.5 > 1 \), the series is convergent.
Thus, Option C is true.
- Analyzing Option D:
The series \( \sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} \frac{1}{n^{1/2}} \) has \( p = 0.5 \le 1 \), so it is a divergent series.
Thus, Option D is false.

Step 3: Final Answer:
The correct statement is $\sum_{\text{n}=1}^{\infty} (\text{a}_\text{n})^3$ is convergent.