Find the radius of convergence of the power series \[ \sum_{n=1}^{\infty} 12^{n} x^{n} n ! \] If needed enter INF for $\infty$. Radius of convergence is
Solution
Step 1 ：We need to calculate the limit as n approaches infinity of the absolute value of the ratio of the (n+1)th term to the nth term of the series. The series is \( \sum_{n=1}^{\infty} 12^{n} x^{n} n ! \), so the nth term is \( 12^{n} x^{n} n ! \) and the (n+1)th term is \( 12^{n+1} x^{n+1} (n+1) ! \).
Step 2 ：Let's calculate the ratio of the (n+1)th term to the nth term: \( \frac{12^{n + 1} x^{n + 1} (n + 1)!}{12^{n} x^{n} n!} \).
Step 3 ：Next, we need to find the limit of this ratio as n approaches infinity. The limit is \( \infty \times \text{sign}(x) \).
Step 4 ：Finally, the radius of convergence is the reciprocal of the limit of the ratio. Since the limit is infinity, the reciprocal is 0.
Step 5 ：Final Answer: The radius of convergence of the power series is \( \boxed{0} \).
