Find the interval of convergence of the power series. (Be sure to include a check for convergence at the endpoints of the interval. If the answer is an interval, enter your answer using interval notation. If the answer is a finite set of values, enter your answers as a comma- separated list of values.) \[ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(x-5)^{n}}{n 7^{n}} \]

Step 1 ：The nth term of the series is given by \(a_n = (-1)^{(n+1)} * (x-5)^n / (n * 7^n)\).

Step 2 ：The (n+1)th term of the series is given by \(a_{(n+1)} = (-1)^{(n+2)} * (x-5)^{(n+1)} / ((n+1) * 7^{(n+1)})\).

Step 3 ：We take the absolute value of the ratio of \(a_{(n+1)}\) to \(a_n\) and take the limit as n approaches infinity: \(\lim_{n \to \infty} |a_{(n+1)} / a_n| = \lim_{n \to \infty} |((-1)^{(n+2)} * (x-5)^{(n+1)} / ((n+1) * 7^{(n+1)})) / ((-1)^{(n+1)} * (x-5)^n / (n * 7^n))|\)

Step 4 ：Simplify the expression to get: \(= \lim_{n \to \infty} |(x-5) / 7 * n / (n+1)|\)

Step 5 ：As n approaches infinity, \(n/(n+1)\) approaches 1, so the limit is \(|(x-5) / 7|\).

Step 6 ：For the series to converge, this limit must be less than 1: \(|(x-5) / 7| < 1\)

Step 7 ：Solving this inequality gives the interval of convergence: \(-7 < x-5 < 7\)

Step 8 ：Adding 5 to all parts of the inequality gives: \(-2 < x < 12\)

Step 9 ：So, the interval of convergence is \((-2, 12)\).

Step 10 ：Now, we need to check the endpoints of the interval, x = -2 and x = 12.

Step 11 ：For x = -2, the series becomes: \(\sum_{n=1}^{\infty} \frac{(-1)^{(n+1)}(-7)^{n}}{n 7^{n}}\). This is an alternating series with terms that decrease in absolute value to 0, so by the Alternating Series Test, it converges.

Step 12 ：For x = 12, the series becomes: \(\sum_{n=1}^{\infty} \frac{(-1)^{(n+1)}7^{n}}{n 7^{n}}\). This simplifies to: \(\sum_{n=1}^{\infty} \frac{(-1)^{(n+1)}}{n}\). This is the alternating harmonic series, which is known to converge.

Step 13 ：Therefore, the series converges at both endpoints, and the interval of convergence is \([-2, 12]\).

Step 14 ：\(\boxed{\text{The interval of convergence is } [-2, 12]}\)