jmc

From the formula for an infinite geometric series, $$\frac{1}{\tau }+\frac{1}{{\tau }^2}+\frac{1}{{\tau }^3}+\cdots =\frac{1/\tau }{1-1/\tau }=\frac{1}{\tau -1}.$$Recall that $\tau$ satisfies ${\tau }^2-\tau -1=0.$ Then $\tau (\tau -1)=1,$ so $$\frac{1}{\tau -1}=\tau .$$Thus, $n=\boxed{1}.$