Question

The series \( \sum_{n=1}^{\infty} \frac{x^n}{n}, (x \gt 0) \) is:

• A. Convergent if \( x \gt 1 \) and divergent if \( x \le 1 \)
• B. Divergent if \( x \lt 1 \) and convergent if \( x \ge 1 \)
• C. Convergent if \( x \lt 1 \) and divergent if \( x \ge 1 \)
• D. Convergent if \( x \lt 1 \) and divergent if \( x \le 1 \)

Hint:

For power series \( \sum a_n x^n \), the radius of convergence \( R \) is \( 1/L \). Always remember to check the convergence at the boundary points \( x = R \) separately.

Answer 1

The Correct Option is C

Concept: The Ratio Test is the primary tool for determining the convergence of power series.

• Let the general term be \( a_n \).

• Calculate the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).

• Convergence occurs if \( L \lt 1 \), divergence if \( L \gt 1 \).

Step 1: Applying the Ratio Test.
Identify \( a_n = \frac{x^n}{n} \) and \( a_{n+1} = \frac{x^{n+1}}{n+1} \). \[ L = \lim_{n \to \infty} \left| \frac{x^{n+1}}{n+1} \cdot \frac{n}{x^n} \right| \] Simplify the expression. \[ L = \lim_{n \to \infty} \left| x \cdot \frac{n}{n+1} \right| = x \cdot \lim_{n \to \infty} \frac{1}{1 + 1/n} \] As \( n \to \infty \), the fraction tends to 1. \[ L = x \cdot 1 = x \]

Step 2: Evaluating the convergence criteria.
The series converges if \( L \lt 1 \), which means \( x \lt 1 \). The series diverges if \( L \gt 1 \), which means \( x \gt 1 \).

Step 3: Testing the boundary point \( x = 1 \).
At \( x = 1 \), the series becomes \( \sum \frac{1}{n} \). This is the Harmonic Series, which is a known divergent p-series. Thus, the series diverges for \( x \ge 1 \).