4.2 For which values of \( x \) will \( \sum_{k=1}^{\infty}(4 x-1)^{k} \)

To determine the values of \( x \) for which the series \( \sum_{k=1}^{\infty}(4 x-1)^{k} \) converges, we can recognize that this is a geometric series. A geometric series converges if the absolute value of the common ratio is less than 1. In this case, the common ratio \( r \) is given by: \[ r = 4x - 1 \] The series converges when: \[ |4x - 1| < 1 \] Now, we can solve this inequality step by step. 1. **Set up the inequality**: \[ -1 < 4x - 1 < 1 \] 2. **Break it into two parts**: - For the left part: \[ -1 < 4x - 1 \] Adding 1 to both sides: \[ 0 < 4x \] Dividing by 4: \[ 0 < x \] - For the right part: \[ 4x - 1 < 1 \] Adding 1 to both sides: \[ 4x < 2 \] Dividing by 4: \[ x < \frac{1}{2} \] 3. **Combine the results**: From the two parts, we have: \[ 0 < x < \frac{1}{2} \] Thus, the series \( \sum_{k=1}^{\infty}(4 x-1)^{k} \) converges for values of \( x \) in the interval: \[ \boxed{(0, \frac{1}{2})} \]