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

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

\[ \sum_{n=1}^{\infty} \frac{\sqrt{n}}{6^n} (x + 4)^n \]

\( R = \)

Find the interval, \( I \), of convergence of the series.

\( I = \)

Solution

To find the radius of convergence, \( R \), use the ratio test. \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\sqrt{n+1}}{6^{n+1}}(x+4)^{n+1} \cdot \frac{6^n}{\sqrt{n}}(x+4)^{-n} \right| & \] \[ = \lim_{n \to \infty} \left| \frac{\sqrt{n+1}}{\sqrt{n}} \cdot \frac{1}{6} \cdot |x+4| \right| & \] \[ = \frac{|x+4|}{6} & \] Set the limit less than 1 for convergence: \[ \frac{|x+4|}{6} < 1 & \] Thus, \[ |x+4| < 6 & \] Therefore, the radius of convergence is: \[ R = 6 & \] To find the interval of convergence, examine the endpoints. At \( x = -10 \): \[ \sum_{n=1}^{\infty} \frac{\sqrt{n}}{6^n}(-6)^n = \sum_{n=1}^{\infty} \sqrt{n} (-1)^n & \] Since \( \sqrt{n} \) does not approach zero, the series diverges. At \( x = 2 \): \[ \sum_{n=1}^{\infty} \frac{\sqrt{n}}{6^n}(6)^n = \sum_{n=1}^{\infty} \sqrt{n} & \] This series also diverges. Thus, the interval of convergence is: \[ I = (-10, 2) & \]