Question

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

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

Hint:

Always remember:
The harmonic series \(\sum \frac{1}{n}\) is the classic boundary case of a divergent series.
Even though the terms \(\frac{1}{n}\) shrink to 0 as \(n \to \infty\), they do not shrink fast enough for the infinite sum to remain finite.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The given series is the famous harmonic series:
\[ \sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots \]
We can determine the convergence of this series using the standard p-series test.
Key Formula or Approach:
The p-series test states that a series of the form \(\sum_{n=1}^{\infty} \frac{1}{n^p}\) is:
- Convergent if \(p > 1\)
- Divergent if \(p \le 1\)

Step 2: Detailed Explanation:
Let us apply the p-series test to the harmonic series:
\[ \sum_{n=1}^{\infty} \frac{1}{n^1} \]
Here, the exponent of \(n\) in the denominator is \(p = 1\).
Since \(p = 1 \le 1\), the series is divergent.
We can also prove this divergence using the integral test:
Let \(f(x) = \frac{1}{x}\). Since \(f(x)\) is continuous, positive, and decreasing on \([1, \infty)\), we can evaluate the improper integral:
\[ \int_{1}^{\infty} \frac{1}{x} \, dx = \lim_{t \to \infty} \left[ \ln|x| \right]_{1}^{t} = \lim_{t \to \infty} (\ln t - \ln 1) = \infty \]
Since the improper integral diverges to infinity, the corresponding series also diverges.
Therefore, the harmonic series is divergent.

Step 3: Final Answer:
The correct option is (B).