Hi! I am currently working on some calc2 online homework problems on Taylor series and Maclaurin series. I have tried a few different answers to this question, but I am really not sure how to go about solving this problem. If anyone can help me I would greatly appreciate it!
Observe that $$\frac{1}{1 - 3x} = \frac{1}{-2 - 3(x - 1)} = -\frac{1}{2 + 3(x - 1)} = -\frac{1/2}{1 + (3/2)(x - 1)}.$$ Therefore $$\frac{1}{1 - 3x} = -\frac{1}{2}\sum_{n = 0}^\infty (-1)^n \left(\frac{3(x - 1)}{2}\right)^n = -\frac{1}{2}\sum_{n = 0}^\infty (-1)^n \left(\frac{3}{2}\right)^n (x - 1)^n, \quad \left|\frac{3(x - 1)}{2}\right| < 1.$$
Because $$\left|\frac{3(x - 1)}{2}\right| < 1 \Longleftrightarrow \frac{1}{3} < x < \frac{5}{3},$$ and the series above diverges when $x = 1/3$ or $x = 5/3$, we conclude that the interval of convergence is $(1/3, 5/3)$.