Question

Let $\sum_{n=1}^\infty \frac{1}{n^2}$ and $\sum_{n=1}^\infty \frac{1}{n^{\frac{1}{2}}}$ be series, then}

• A. both are convergent
• B. both are divergent
• C. $\sum_{n=1}^\infty \frac{1}{n^2}$ is convergent and $\sum_{n=1}^\infty \frac{1}{n^{\frac{1}{2}}}$ is divergent
• D. $\sum_{n=1}^\infty \frac{1}{n^2}$ is divergent and $\sum_{n=1}^\infty \frac{1}{n^{\frac{1}{2}}}$ is convergent

Hint:

Remember the threshold value for the p-series test:
- Any power \(\mathbf{p}\) strictly greater than 1 converges.
- Even a value like \(p = 1.0001\) converges.
- Any power \(\mathbf{p}\) less than or equal to 1 (including the harmonic series where \(p=1\)) diverges.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We evaluate the convergence of these infinite series using the p-series test.
The p-series test is a standard convergence test for positive term series of a specific power form.

Step 2: Key Formula or Approach:
The standard p-series is written as:
\[ \sum_{n=1}^\infty \frac{1}{n^p} \] - The series converges if \(p > 1\).
- The series diverges if \(p \le 1\).

Step 3: Detailed Explanation:
Let us apply the p-series test to both given series:
1. Analyzing the first series:
\[ S_1 = \sum_{n=1}^\infty \frac{1}{n^2} \] - Here, the exponent is \(p = 2\).
- Since \(2 > 1\), this series is convergent.
2. Analyzing the second series:
\[ S_2 = \sum_{n=1}^\infty \frac{1}{n^{\frac{1}{2}}} \] - Here, the exponent is \(p = \frac{1}{2}\) (or 0.5).
- Since \(\frac{1}{2} \le 1\), this series is divergent.
Therefore, the first series is convergent and the second series is divergent.

Step 4: Final Answer:
The correct statement is that \(\sum \frac{1}{n^2}\) is convergent and \(\sum \frac{1}{n^{\frac{1}{2}}}\) is divergent, matching Option (C).