Convergence of geometric series of probabilities.

Let X be an absolutely continuous random variable with strictly positive pdf f and cdf F. Moreover, let $\gamma \in (0,1)$. I am interested when $\sum _{n=1}^{\mathrm{\infty}}P(|X|<{\gamma}^{n})=\sum _{n=1}^{\mathrm{\infty}}F({\gamma}^{n})-F(-{\gamma}^{n})$ converges. One option would be to impose that F is K-Lipschitz continuous, since then $F({\gamma}^{n})-F(-{\gamma}^{n})\le 2K{\gamma}^{n}$ and thus the series reduces to a geometric one. Is it also sufficient if F is assumed to be uniformly continuous? If not, are there additional assumptions that do ensure this is enough?

Let X be an absolutely continuous random variable with strictly positive pdf f and cdf F. Moreover, let $\gamma \in (0,1)$. I am interested when $\sum _{n=1}^{\mathrm{\infty}}P(|X|<{\gamma}^{n})=\sum _{n=1}^{\mathrm{\infty}}F({\gamma}^{n})-F(-{\gamma}^{n})$ converges. One option would be to impose that F is K-Lipschitz continuous, since then $F({\gamma}^{n})-F(-{\gamma}^{n})\le 2K{\gamma}^{n}$ and thus the series reduces to a geometric one. Is it also sufficient if F is assumed to be uniformly continuous? If not, are there additional assumptions that do ensure this is enough?