How can I obtain a closed form for the mean of the following discreet distribution?
$$ \Pr(X=k) = \begin{cases} p\,(1 - p)^k & \text{, if $k < n$} \\ p\,(1 - p)^n\,(1 - q)^{(k-n)}&, \text{ if $k \ge n$} \\ \end{cases}, $$
where $p$, $q$ and $n$ are parameters of the discreet distribution. Note that, if $p=q$ or $n = 0$ or $n\rightarrow\infty$, then the distribution becomes the geometric distribution.
This appears to be a computational exercise.
Let $$G(x)=\sum_{k=0}^{n-1}x^kp(1-p)^k+\sum_{k=n}^\infty x^kp(1-p)^n(1-q)^{k-n}$$ be the probability generating function.
Both parts are computable in closed form using geometric series, and then you can differentiate it to get the mean.