I claim that 
$$ \sum_{k=0}^\infty\mu^k(1-\mu^kt)^n\asymp_\mu\frac{\log(nt)}{nt}. $$
For the lower bound observe that $\frac{1}{nt}<\mu^k<\frac{2}{nt}$ holds in a range $k\asymp_\mu\log(nt)$ and any such $k$ contributes $\gg\frac{1}{nt}$ to the sum. For the upper bound split and estimate the sum as follows:
$$ \sum_{k=0}^\infty\mu^k(1-\mu^kt)^n = \sum_{k:\mu^k<\lambda}\dots + \sum_{k:\mu^k\geq\lambda}\dots < \frac{\lambda}{1-\mu}+\frac{(1-\lambda t)^n}{1-\mu}\ll_\mu \lambda+e^{-\lambda nt}.$$
For $\lambda:=\frac{\log(nt)}{nt}$ the right hand side is $\ll\frac{\log(nt)}{nt}$, so we are done.

Note: The above bounds show that the $O(1/n)$ conjecture fails for any $0 < t < 1$.