Calculating Radius of Convergence

We need to find the radius of convergence $R$ for the power series: $ s = \sum_{m=1}^{\infty} \frac{m^2}{3^m} (x - 2)^m $ This is a power series of the form $\sum c_m (x-a)^m$, where $a = 2$ and the coefficients are $c_m = \frac{m^2}{3^m}$.

Applying the Ratio Test

We use the Ratio Test for power series, which relies on the limit of the ratio of consecutive coefficients. The radius of convergence $R$ is given by: $ R = \frac{1}{\lim_{m \to \infty} \left| \frac{c_{m+1}}{c_m} \right|} $

First, let's find the ratio $\frac{c_{m+1}}{c_m}$: $ c_{m+1} = \frac{(m+1)^2}{3^{m+1}} $ $ \frac{c_{m+1}}{c_m} = \frac{\frac{(m+1)^2}{3^{m+1}}}{\frac{m^2}{3^m}} = \frac{(m+1)^2}{3^{m+1}} \cdot \frac{3^m}{m^2} $ $ = \frac{(m+1)^2}{m^2} \cdot \frac{3^m}{3^{m+1}} = \left(\frac{m+1}{m}\right)^2 \cdot \frac{1}{3} $ $ = \left(1 + \frac{1}{m}\right)^2 \cdot \frac{1}{3} $

Determining the Limit

Now, we compute the limit as $m$ approaches infinity: $ \lim_{m \to \infty} \left| \frac{c_{m+1}}{c_m} \right| = \lim_{m \to \infty} \left| \left(1 + \frac{1}{m}\right)^2 \cdot \frac{1}{3} \right| $ As $m \to \infty$, $\frac{1}{m} \to 0$. Therefore, the limit becomes: $ = \left(1 + 0\right)^2 \cdot \frac{1}{3} = 1^2 \cdot \frac{1}{3} = \frac{1}{3} $

Calculating the Radius R

Using the formula for the radius of convergence: $ R = \frac{1}{\lim_{m \to \infty} \left| \frac{c_{m+1}}{c_m} \right|} = \frac{1}{1/3} $ $ R = 3 $ The series converges for values of $x$ such that $|x-2| < R$, which means $|x-2| < 3$. The question asks for the value $R$ given $|x-2| \le R$. Based on the ratio test calculation, the radius of convergence is $R=3$.