5.1 Given: \( \sum_{n=1}^{\infty}\left(x^{2}-3 x+1\right)^{n} \) Determine for which values of \( x \) the series will be convergent.

To determine for which values of \( x \) the series \( \sum_{n=1}^{\infty}\left(x^{2}-3 x+1\right)^{n} \) converges, we can recognize that this is a geometric series with the common ratio \( r = x^{2} - 3x + 1 \). A geometric series converges when the absolute value of the common ratio is less than 1. Thus, we require: \[ |x^{2} - 3x + 1| < 1 \] This leads us to two inequalities: \[ -1 < x^{2} - 3x + 1 < 1 \] We can solve each part of this compound inequality separately. First, solving: 1. \( x^{2} - 3x + 1 > -1 \) This simplifies to: \[ x^{2} - 3x + 2 > 0 \] Factoring gives: \[ (x-1)(x-2) > 0 \] This inequality holds for \( x < 1 \) or \( x > 2 \). 2. Now, solving \( x^{2} - 3x + 1 < 1 \) leads to: \[ x^{2} - 3x < 0 \] Factoring yields: \[ x(x-3) < 0 \] This inequality holds for \( 0 < x < 3 \). Now, combining both solutions, we look for the intersection of \( x < 1 \) or \( x > 2 \) with \( 0 < x < 3 \). The values of \( x \) for convergence are: \[ (0, 1) \cup (2, 3) \] Thus, the series converges for \( x \) in the intervals \( (0, 1) \) and \( (2, 3) \).