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 $$ \sum_{n=1}^{\infty}P(|X|<\gamma^n) = \sum_{n=1}^{\infty} F(\gamma^n) - F(-\gamma^n) $$ 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?
Thank you in advance!
Uniform continuity is insufficient. Let $$F(x) = \begin{cases} 0&x\le -e^{-2}\\ 1/2 - 1/\ln(-1/x)&-e^{-2}\le x\le 0\\ 1/2 + 1/\ln(1/x)&0\le x\le e^{-2}\\ 1&x\ge e^{-2} \end{cases}$$ This is a cdf which is uniformly continuous and for which $\sum_{n}F(\gamma^n)$ does not converge for any $0<\gamma<1$.
For additional assumptions, it suffices for $F$ to be $\alpha$-Hölder continuous at zero for some $\alpha>0$. So if $F(x)=1/2+\text{sign}(x)\sqrt{|x|}$ in a neighborhood of $0$, then your sum converges, even though $F$ is not Lipschitz.