 Solution

Normalization of a probability distribution and the geometric series give $1=A/(1-\rho )$, so $A=1-\rho$. For a nonnegative integer $n$, the strict tail starts at $n+1$:For nonnegative integers $n>m$, the event $\{N>n\}$ is contained in $\{N>m\}$. ThereforeThis is the memorylessness of the geometric distribution with the zero-based counting convention. Keeping strict versus non-strict inequalities explicit is essential: $N-m-1$, conditional on $N>m$, has the original distribution of $N$.