Question

For what value of \(p\), the series \(\sum_{n=1}^\infty \frac{1}{n^p}\) is convergent

• A. 2
• B. 1
• C. \(\frac{1}{2}\)
• D. \(\frac{2}{3}\)

Hint:

The p-series test is a very simple and important rule to memorize:
\[ \sum \frac{1}{n^p} \text{ converges } \iff p > 1 \] This rule helps you quickly find bounding series when using the comparison test.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This question concerns the convergence of an infinite series of positive terms.
Specifically, it tests the p-series test (also known as the hyper-harmonic series test).

Step 2: Detailed Explanation:
The general form of a p-series is:
\[ \sum_{n=1}^\infty \frac{1}{n^p} = 1 + \frac{1}{2^p} + \frac{1}{3^p} + \frac{1}{4^p} + \dots \] According to the p-series convergence theorem (which is proved using the integral test):
- The series converges if and only if \( p > 1 \).
- The series diverges if \( p \le 1 \).
Let us evaluate each of the given options against this mathematical rule:
- Option (A): \( p = 2 \)
Since \( 2 > 1 \), the series \( \sum_{n=1}^\infty \frac{1}{n^2} \) satisfies the condition for convergence.
This series is famous in mathematics (related to the Basel problem) and converges to \( \frac{\pi^2}{6} \).
Thus, this option is correct.
- Option (B): \( p = 1 \)
The series becomes \( \sum_{n=1}^\infty \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \dots \), which is the classical harmonic series. Since \( 1 \le 1 \), this series diverges.
- Option (C): \( p = 1/2 \)
The series is \( \sum_{n=1}^\infty \frac{1}{\sqrt{n}} \). Since \( 1/2 \le 1 \), this series diverges.
- Option (D): \( p = 2/3 \)
Since \( 2/3 \le 1 \), this series diverges.
Therefore, the series is convergent only when \( p = 2 \).

Step 3: Final Answer:
The series is convergent for \( p = 2 \).
Therefore, the correct choice is Option (A).