May someone check my work?

$\displaystyle \sum_{n}^{\infty}(n3^{-n})(x-2)^n$

To find R

$\displaystyle \lim_{n \rightarrow \infty} \left| \frac{n3^{-n}}{(n+1)3^{1-n}} \right| = \frac{1}{3}\lim_{n \rightarrow \infty} \left| \frac{n}{n+1} \right|=\frac{1}{3}$

To find the interval of convergence, C

$\displaystyle \left| x-2 \right|<\frac{1}{3} \Rightarrow \frac{5}{3}<x<\frac{7}{3}$

For $\displaystyle x=\frac{5}{3}$:

$\displaystyle \sum_{n}^{\infty}(n3^{-n})\left\(\frac{-1}{3}\right\)^n=\sum_{n}^{\infty}n(-1)^n\left\(\frac{1}{3}\right\)^{2n}$ which converges by alternating series test.

For $\displaystyle x=\frac{7}{3}$:

$\displaystyle \sum_{n}^{\infty}(n3^{-n})\left\(\frac{1}{3}\right\)^n=\sum_{n}^{\infty} \frac{n}{9^n}$ which converges by the root test.

So C=$\displaystyle \frac{5}{3}\leq x \leq \frac{7}{3}\right]$