Question

The series \(\sum_{n=1}^{\infty} \frac{1}{n}\) is :

• A. Convergent
• B. Divergent
• C. Absolutely Convergent
• D. Conditionally Convergent

Hint:

The harmonic series is a classic boundary case of a divergent series:
Even though the terms \(\frac{1}{n} \to 0\) as \(n \to \infty\) (satisfying the divergence test's necessary condition), they do not decrease fast enough to yield a finite sum.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This question asks about the convergence or divergence of the standard harmonic series:
\[ \sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots \]
Key Formula or Approach:
We use the \(p\)-series test. A \(p\)-series of the form:
\[ \sum_{n=1}^{\infty} \frac{1}{n^p} \] - Converges if \(p > 1\).
- Diverges if \(p \le 1\).

Step 2: Detailed Explanation:

• Let us analyze the given series:
\[ \sum_{n=1}^{\infty} \frac{1}{n} = \sum_{n=1}^{\infty} \frac{1}{n^1} \]

• Here, the power \(p\) is equal to 1.

• Applying the \(p\)-series test:
Since \(p = 1\), the condition \(p \le 1\) is satisfied, which means the series diverges.

• We can also prove this using the integral test:
Evaluate the improper integral:
\[ \int_{1}^{\infty} \frac{1}{x} \, dx = \lim_{t \to \infty} [\ln(x)]_1^t = \lim_{t \to \infty} \ln(t) - 0 = \infty \] Since the improper integral diverges, the series also diverges.

Step 3: Final Answer:
The harmonic series \(\sum_{n=1}^{\infty} \frac{1}{n}\) is divergent.