I though to apply Borel-Cantelli, however it does not seem to work. m and $\mu$ are constants, with $0<m<\mu$, with random identically distributed variables $Z_{n}$, which take values in $\mathbb{N}_{0}$. What I got was (all of them are nonnegativ so I leave the abs away):\ $\sum^{\infty}_{n=0}\mathbb{P}(\frac{Z_{n}}{\mu^{n}}>\epsilon) =\sum^{\infty}_{n=0}\mathbb{P}(Z_{n}>\mu^{n}\epsilon)=\sum^{\infty}_{n=0}\mathbb{E}(\mathbb{1}_{\{Z_{n}>\mu^{n}\epsilon\}})=\mathbb{E}(\sum^{\infty}_{n=0}\mathbb{1}_{\{Z_{n}>\mu^{n}\epsilon\}})$ \ I though when I get expected value it would be easier to estimate because of linearity and at this point I tried to estimate the last part with $\sum^{\infty}_{n=0}\mathbb{P}(|\frac{Z_{n}}{m^{n}}-W|>\epsilon)$ turning it into expected value. However it doesn´t seem to help a lot.
Well now because of the comment I notice, that is sufficient to use , that $\frac{m}{\mu}<1$ then because W is some finite value $\lim_{n\to \infty}\frac{Z_{n}}{\mu^{n}}=\lim_{n\to \infty}\frac{Z_{n}}{\mu^{n}}\frac{m^{n}}{m^{n}}=W\lim_{n\to \infty}(\frac{m}{\mu})^{n}=0$