Question

The series \(\displaystyle\sum_{n=0}^{\infty}(2x)^n\) converges if ____.

• A. \(-1\leq x\leq 1\)
• B. \(-\dfrac{1}{2}<x<\dfrac{1}{2}\)
• C. \(-2<x<2\)
• D. \(-\dfrac{1}{2}\leq x\leq\dfrac{1}{2}\)

Hint:

For a geometric series \(\sum r^n\), convergence occurs only when \(|r|<1\). Always check the end points separately.

Answer 1

The Correct Option is B

Concept:
The given series is: \[ \sum_{n=0}^{\infty}(2x)^n \] This is a geometric series. The general form of a geometric series is: \[ \sum_{n=0}^{\infty}r^n \] A geometric series converges only when: \[ |r|<1 \] and diverges when: \[ |r|\geq 1 \]

Step 1: Identify the common ratio.
Given series: \[ \sum_{n=0}^{\infty}(2x)^n \] Here, the common ratio is: \[ r=2x \]

Step 2: Apply convergence condition.
For convergence of geometric series: \[ |r|<1 \] Substitute: \[ r=2x \] \[ |2x|<1 \]

Step 3: Solve the inequality.
\[ |2x|<1 \] This gives: \[ -1<2x<1 \] Now divide throughout by \(2\): \[ -\frac{1}{2}<x<\frac{1}{2} \]

Step 4: Check the end points.
At: \[ x=\frac{1}{2} \] we get: \[ 2x=1 \] So the series becomes: \[ \sum_{n=0}^{\infty}1^n \] \[ =1+1+1+\cdots \] This diverges. At: \[ x=-\frac{1}{2} \] we get: \[ 2x=-1 \] So the series becomes: \[ \sum_{n=0}^{\infty}(-1)^n \] \[ =1-1+1-1+\cdots \] This also does not converge in the usual sense.

Step 5: Final conclusion.
Therefore, the series converges only for: \[ -\frac{1}{2}<x<\frac{1}{2} \] \[ \therefore \text{Correct Answer is (B)} \]