What is the difference between these two Laurent series expansions

Question

I am working through problems in "Cracking the GRE Mathematics Subject Test" (4th edition) and one of the practice problems asks for the coefficient of $(z - 2)^{-1}$ in the Laurent series for $f(z) = \frac{1}{z-5}$ centered at $z = 2$. For my solution I did $$\frac{1}{z-5} = -\frac{1}{3 - (z - 2)} = -\frac{1}{3}\frac{1}{1 - \frac{z - 2}{3}} = \sum_{n = 0}^\infty{3^{-n - 1}(z - 2)^n}.$$ This expansion is valid on $|z - 2| < 3$. Hence, the coefficent is $0$. However, for the solution in the book they say $$\frac{1}{z-5} = \frac{\frac{1}{z - 2}}{1 - \frac{3}{z - 2}} = \frac{1}{z-2}\sum_{n = 0}^\infty{\left(\frac{3}{z-2}\right)^n} = \sum_{n = 0}^\infty{3^n(z-2)^{-n - 1}}.$$ This expansion is valid on $|z - 2| > 3$. Hence, the coefficent, using this expansion, is 1. Is there a reason that the second one is correct and the first one is not?