Power Series Radius of Convergence Calculation

To determine the radius of convergence (R) for the power series $ \sum_{n=0}^{\infty} \frac{(x-3)^n}{3^n n!} $, we apply the Ratio Test.

Ratio Test Application

The Ratio Test is used for power series of the form $ \sum a_n (x-c)^n $. We examine the limit $ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| $, where $ a_n $ are the coefficients of the series terms excluding $ (x-c)^n $. The radius of convergence R is given by $ R = \frac{1}{L} $.

For the series $ \sum_{n=0}^{\infty} \frac{(x-3)^n}{3^n n!} $, the coefficients are $ c_n = \frac{1}{3^n n!} $. Thus, $ c_{n+1} = \frac{1}{3^{n+1} (n+1)!} $.

Calculate the limit L:

$ L = \lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{1}{3^{n+1} (n+1)!}}{\frac{1}{3^n n!}} \right| $

Simplify the ratio:

$ L = \lim_{n \to \infty} \left| \frac{3^n n!}{3^{n+1} (n+1)!} \right| = \lim_{n \to \infty} \left| \frac{3^n \cdot n!}{3 \cdot 3^n \cdot (n+1) \cdot n!} \right| $

$ L = \lim_{n \to \infty} \left| \frac{1}{3(n+1)} \right| $

Evaluate the limit:

$ L = \frac{1}{3} \lim_{n \to \infty} \frac{1}{n+1} = \frac{1}{3} \cdot 0 = 0 $

Radius of Convergence Result

Since $ L = 0 $, the radius of convergence is:

$ \frac{1}{R} = 0 $

$ R = \infty $

Final Answer

The radius of convergence for the given power series is $ \infty $.