\( \sum _ { n = 0 } ^ { \infty } n ! ( 2 x - 1 ) ^ { n } \)

To evaluate the series \( \sum_{n=0}^{\infty} n! (2x - 1)^n \), we can recognize that the series resembles a known function. Specifically, we can utilize the exponential generating function for the factorial. The series can be rewritten as follows: \[ \sum_{n=0}^{\infty} n! y^n = \frac{1}{1 - y} \] for \( |y| < 1 \). In our case, let \( y = 2x - 1 \). Therefore, the series converges for \( |2x - 1| < 1 \), or \( 0 < x < 1 \). So we have: \[ \sum_{n=0}^{\infty} n! (2x - 1)^n = e^{(2x - 1)} \] However, this is just a transformation, and we need to handle the sum properly: To find the function corresponding to the series, if we take the sum \( \sum_{n=0}^{\infty} \frac{(2x - 1)^n}{n!} \), it yields \( e^{2x-1} \), indicating that our original series diverges for most values. To find the proper sum expansion, we'd investigate methods like using integral transforms or addressing convergence regions. While we present this, proper techniques like power series and complex analysis might give cleaner conclusions about evaluating convergence under real conditions, leading us to realize higher methodologies can yield cleaner results. In any case, the series diverges for most values, leading to fascinating behavior under constraints and functional analysis in complex variables!