find the interval of convergence \( \sum_{n=0}^{\infty} \frac{(x-11)^{n-1}}{11^{n-1}} \)

To determine the interval of convergence, we can first rewrite the series: \[ \sum_{n=0}^{\infty} \frac{(x-11)^{n-1}}{11^{n-1}} = \sum_{n=1}^{\infty} \frac{(x-11)^{n-1}}{11^{n-1}} \quad (\text{shifting index}) \] This transformation leads to: \[ \sum_{n=1}^{\infty} \left(\frac{x-11}{11}\right)^{n-1} = \sum_{n=0}^{\infty} \left(\frac{x-11}{11}\right)^{n} \] This is a geometric series with the ratio \( r = \frac{x - 11}{11} \). A geometric series converges if the absolute value of the ratio is less than 1, which gives us: \[ \left|\frac{x-11}{11}\right| < 1 \] This inequality can be solved as follows: 1. Expand the absolute value inequality: \[ -1 < \frac{x-11}{11} < 1 \] 2. Multiply through by 11 (which is positive and maintains the inequalities): \[ -11 < x - 11 < 11 \] 3. Add 11 to all parts of the inequality: \[ 0 < x < 22 \] Thus, the interval of convergence is \( (0, 22) \). Finally, we should check the endpoints \( x = 0 \) and \( x = 22 \) to ascertain whether they should be included in the interval: - For \( x = 0 \): \[ \sum_{n=0}^{\infty} \frac{(0-11)^{n-1}}{11^{n-1}} = \sum_{n=0}^{\infty} (-1)^{n-1} \] This series does not converge. - For \( x = 22 \): \[ \sum_{n=0}^{\infty} \frac{(22-11)^{n-1}}{11^{n-1}} = \sum_{n=0}^{\infty} 1 \] This series also does not converge. Thus, the interval of convergence is confirmed to be \( (0, 22) \).