Clearly, the possible common ratios are the 2013 roots $r_{1}, r_{2}, \ldots, r_{2013}$ of the equation $r^{2013}=\frac{1}{2013^{2013}}$. We want the sum of the values of $x_{n}=\frac{1}{1-r_{n}}$, so we consider the polynomial whose roots are $x_{1}, x_{2}, \ldots, x_{2013}$. It is easy to see that $\left(1-\frac{1}{x_{n}}\right)^{2013}=\frac{1}{2013^{2013}}$, so it follows that the $x_{n}$ are the roots of the polynomial equation $\frac{1}{2013^{2013}} x^{2013}-(x-1)^{2013}=0$. The leading coefficient of this polynomial is $\frac{1}{2013^{2013}}-1$, and it follows easily from the Binomial Theorem that the next coefficient is 2013, so our answer is, by Vieta's Formulae, $$-\frac{2013}{\frac{1}{2013^{2013}}-1}=\frac{2013^{2014}}{2013^{2013}-1}$$