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 interval of convergence is an interval, enter your answer using interval notation. If the interval of convergence is a finite set, enter your answer using set notation.)
\[
\sum_{n=0}^{\infty}(-1)^{n+1}(n+5) x^{n}
\]

Step 1 ：The nth term of the series is \(a_n = (-1)^{(n+1)}(n+5)x^n\). The (n+1)th term is \(a_{(n+1)} = (-1)^{(n+2)}((n+1)+5)x^{(n+1)}\).

Step 2 ：We find the ratio of \(a_{(n+1)}\) to \(a_n\): \(a_{(n+1)} / a_n = [(-1)^{(n+2)}((n+1)+5)x^{(n+1)}] / [(-1)^{(n+1)}(n+5)x^n] = [(-1)(n+6)x] / [(n+5)] = -x(n+6)/(n+5)\).

Step 3 ：We take the absolute value and the limit as n approaches infinity: \(\lim_{n \to \infty} |a_{(n+1)} / a_n| = \lim_{n \to \infty} |-x(n+6)/(n+5)| = |x| \lim_{n \to \infty} |(n+6)/(n+5)| = |x|\).

Step 4 ：The series converges if this limit is less than 1, i.e., if \(|x| < 1\). This gives us the interval \((-1, 1)\).

Step 5 ：Now we check the endpoints. If x = -1, the series becomes: \(\sum_{n=0}^{\infty}(-1)^{n+1}(n+5)(-1)^{n}\). This is an alternating series with terms decreasing in absolute value, so by the Alternating Series Test, it converges.

Step 6 ：If x = 1, the series becomes: \(\sum_{n=0}^{\infty}(-1)^{n+1}(n+5)\). This is also an alternating series with terms decreasing in absolute value, so it also converges.

Step 7 ：Therefore, the interval of convergence is \([-1, 1]\).