Harvard-MIT November Tournament

Solution:
Notice that the series is geometric with ratio $x$, so it converges if $-1 < x < 1$. Also notice that where $g_{1}(x)$ is defined, it is equal to $\frac{1}{3(1-x)}$. The image of $g_{1}(x)$ is then the interval $\left(\frac{1}{6}, \infty\right)$. The image of $g_{2}(x)$ is simply the values of $g_{1}(x)$ for $x$ in $\left(\frac{1}{6}, 1\right)$, which is the interval $\left(\frac{2}{5}, \infty\right)$. Similarly, the image of $g_{3}(x)$ is $\left(\frac{5}{9}, \infty\right)$, the image of $g_{4}(x)$ is $\left(\frac{3}{4}, \infty\right)$, and the image of $g_{5}(x)$ is $\left(\frac{4}{3}, \infty\right)$. As this does not intersect the interval $(-1, 1)$, $g_{6}(x)$ is not defined for any $x$, so the answer is $5$.