Question: Find the radius of convergence, \( R \), of the series. \[ \sum_{n=0}^{\infty}…

Find the radius of convergence, \( R \), of the series.

\[ \sum_{n=0}^{\infty} \frac{(x-4)^n}{n^3 + 1} \]

\( R = \boxed{\phantom{0}} \)

Find the interval, \( I \), of convergence of the series. (Enter your answer using interval notation.)

\( I = \boxed{\phantom{0}} \)

Solution

To find the radius of convergence \( R \) and the interval of convergence \( I \) for the series \[ \sum_{n=0}^{\infty} \frac{(x-4)^n}{n^3 + 1} \] Step 1: Apply the Root Test to determine the radius of convergence. \[ L = \limsup_{n \to \infty} \left| \frac{(x-4)^n}{n^3 + 1} \right|^{1/n} = |x - 4| \cdot \limsup_{n \to \infty} \left( \frac{1}{n^3 + 1} \right)^{1/n} \] Step 2: Evaluate the limit superior. \[ \limsup_{n \to \infty} \left( \frac{1}{n^3 + 1} \right)^{1/n} = 1 \] Step 3: Set the limit \( L \) less than 1 for convergence. \[ |x - 4| < 1 \] Step 4: Determine the radius of convergence \( R \). \[ R = 1 \] Step 5: Find the interval of convergence by solving the inequality. \[ -1 < x - 4 < 1 \\ 3 < x < 5 \] Step 6: Check convergence at the endpoints \( x = 3 \) and \( x = 5 \). At \( x = 3 \): \[ \sum_{n=0}^{\infty} \frac{(-1)^n}{n^3 + 1} \] This is an alternating series with terms decreasing in magnitude to 0, so it converges. At \( x = 5 \): \[ \sum_{n=0}^{\infty} \frac{1}{n^3 + 1} \] This is a p-series with \( p = 3 > 1 \), so it converges. Final Answers: \[ R = 1 \] \[ I = [3, 5] \]