# 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?

Convergence of geometric series of probabilities.
Let X be an absolutely continuous random variable with strictly positive pdf f and cdf F. Moreover, let $\gamma \in \left(0,1\right)$. I am interested when $\sum _{n=1}^{\mathrm{\infty }}P\left(|X|<{\gamma }^{n}\right)=\sum _{n=1}^{\mathrm{\infty }}F\left({\gamma }^{n}\right)-F\left(-{\gamma }^{n}\right)$ converges. One option would be to impose that F is K-Lipschitz continuous, since then $F\left({\gamma }^{n}\right)-F\left(-{\gamma }^{n}\right)\le 2K{\gamma }^{n}$ 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?
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Step 1
Uniform continuity is insufficient. Let $F\left(x\right)=\left\{\begin{array}{ll}0& x\le -{e}^{-2}\\ 1/2-1/\mathrm{ln}\left(-1/x\right)& -{e}^{-2}\le x\le 0\\ 1/2+1/\mathrm{ln}\left(1/x\right)& 0\le x\le {e}^{-2}\\ 1& x\ge {e}^{-2}\end{array}$.
Step 2
This is a cdf which is uniformly continuous and for which $\sum _{n}F\left({\gamma }^{n}\right)$ 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 $F\left(x\right)=1/2+\text{sign}\left(x\right)\sqrt{|x|}$ in a neighborhood of 0, then your sum converges, even though F is not Lipschitz.