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

Step 1: Apply the root test instead of the ratio test.
For \( \sum \dfrac{x^n}{n} \), the root test looks at \( \lim_{n\to\infty} \left( \dfrac{x^n}{n} \right)^{1/n} = x \cdot \lim_{n\to\infty} n^{-1/n} \). Since \( n^{1/n} \to 1 \) as \( n \to \infty \), this limit is simply \( x \).

Step 2: Use the root test conclusion.
The series converges when this limit is less than 1, so it converges for \( x \lt 1 \), and diverges for \( x \gt 1 \).

Step 3: Check the boundary \( x=1 \) separately.
At \( x=1 \), the series is \( \sum \dfrac{1}{n} \), the harmonic series, which diverges. So the series converges only for \( x \lt 1 \) and diverges for \( x \ge 1 \), matching option (C).