Question: Find the radius of convergence, \( R \), of the series. \[ \sum_{n=1}^{\infty} \frac{x^n}{3…

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

\[ \sum_{n=1}^{\infty} \frac{x^n}{3 \cdot 7 \cdot 11 \cdots (4n-1)} \]

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

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: \[ \sum_{n=1}^{\infty} \frac{x^n}{3 \cdot 7 \cdot 11 \cdot \ldots \cdot (4n-1)} \] Step 1: Apply the Ratio Test \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \] Step 2: Express \( a_{n} \) and \( a_{n+1} \) \[ a_n = \frac{x^n}{3 \cdot 7 \cdot 11 \cdot \ldots \cdot (4n-1)} \] \[ a_{n+1} = \frac{x^{n+1}}{3 \cdot 7 \cdot 11 \cdot \ldots \cdot (4(n+1)-1)} = \frac{x^{n+1}}{3 \cdot 7 \cdot 11 \cdot \ldots \cdot (4n+3)} \] Step 3: Form the Ratio \( \frac{a_{n+1}}{a_n} \) \[ \frac{a_{n+1}}{a_n} = \frac{x^{n+1}}{3 \cdot 7 \cdot 11 \cdot \ldots \cdot (4n+3)} \times \frac{3 \cdot 7 \cdot 11 \cdot \ldots \cdot (4n-1)}{x^n} = \frac{x}{4n + 3} \] Step 4: Compute the Limit \[ \lim_{n \to \infty} \left| \frac{x}{4n + 3} \right| = 0 \] Step 5: Determine the Radius of Convergence Since the limit is 0 for all real numbers \( x \), the radius of convergence \( R \) is: \[ R = \infty \] Step 6: Determine the Interval of Convergence Since \( R = \infty \), the interval of convergence \( I \) is all real numbers: \[ I = (-\infty, \infty) \] Final Answers: Radius of convergence, \( R \), is: \[ R = \infty \] Interval of convergence, \( I \), is: \[ I = (-\infty, \infty) \]