\(\displaystyle \sum_{n=0}^{\infty}\frac{3^{n}}{x^{n}}\)
Try the ratio test for absolute convergence.
\(\displaystyle {\rho}=\lim_{n\to {\infty}}\left|\frac{u_{k+1}}{u_{k}}\right|=\lim_{n\to {\infty}}\left|\frac{3^{n+1}}{x^{n+1}}\cdot\frac{x^{n}}{3^{n}}\right|=\lim_{n\to \infty}\left|\frac{3}{x}\right|=\left|\frac{3}{x}\right|\)
The ratio test for absolute convergence implies that the series converges absolutely if \(\displaystyle {\rho}=\left|\frac{3}{x}\right|<1\) and
diverges if \(\displaystyle {\rho}=\left|\frac{3}{x}\right|>1\)
Thus, the series converges if \(\displaystyle (-\infty, \;\ -3)\cup (3, \;\ \infty)\)
