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 the values of \( x \) for which the series \[ \sum_{n=1}^{\infty}\left(x^{2}-3 x+1\right)^{n} \] converges, we can analyze it as a geometric series. ### Step 1: Recognize the Series as Geometric The given series is a geometric series where each term is of the form \( r^n \) with \( r = x^2 - 3x + 1 \). A geometric series \[ \sum_{n=1}^{\infty} r^n \] converges if and only if the absolute value of the common ratio \( r \) satisfies: \[ |r| < 1 \] ### Step 2: Apply the Convergence Condition For our series, the condition for convergence becomes: \[ |x^2 - 3x + 1| < 1 \] This inequality can be broken down into two separate inequalities: \[ -1 < x^2 - 3x + 1 < 1 \] ### Step 3: Solve the Inequalities **First Inequality:** \[ x^2 - 3x + 1 < 1 \] Simplify: \[ x^2 - 3x < 0 \] Factor: \[ x(x - 3) < 0 \] This inequality holds when \( x \) is between 0 and 3: \[ 0 < x < 3 \] **Second Inequality:** \[ x^2 - 3x + 1 > -1 \] Simplify: \[ x^2 - 3x + 2 > 0 \] Factor: \[ (x - 1)(x - 2) > 0 \] This inequality holds when \( x < 1 \) or \( x > 2 \). ### Step 4: Find the Intersection of Solutions Combine the results from both inequalities: \[ 0 < x < 3 \quad \text{and} \quad (x < 1 \text{ or } x > 2) \] This intersection yields: \[ 0 < x < 1 \quad \text{or} \quad 2 < x < 3 \] ### Final Answer The series is convergent for all real numbers \( x \) such that \( 0 < x < 1 \) or \( 2 < x < 3 \). \[ \boxed{\,0 < x < 1\quad\text{or}\quad2 < x < 3\,} \]