Finding marginal CDF from a joint PDF and CDF 
The number of users logged onto a system, N and the time T until the next user >logs off have joint probability given by:
$$P(N=n,X\leq t)=(1-p)p^{n-1}(1-e^{-n\lambda t}), n=1,2,\dots, t>0$$
Find the marginal PMF of $N$ and the marginal CDF of $X$.

I'm stuck at finding the CDF of $X$. To find the PMF of $N$, I simply find $P(N=n, X\leq \infty)$ right? And to find the CDF of $X$, I have to find $P(N=n, X \leq t)$ for all n right?
So since $n$ is discrete,
$$\sum_{n=1}^{\infty}(1-p)p^{n-1}(1-e^{-n\lambda t})=\frac{1-p}{p}\sum_{n=1}^{\infty}p^n(1-e^{-n\lambda t})$$
But I have no idea how to continue on from here. Unless I'm approaching the question wrongly?
 A: Standard way (assuming throughout that $0\le p<1$):
\begin{align*}
P(X\le t)&=\sum_{n=1}^{\infty}P(N=n, X\le t)=\frac{1-p}p\sum_{n=1}^{\infty}p^n(1-e^{-n\lambda t})\\&=\frac{1-p}{p}\left(\sum_{n=1}^{\infty}p^{n}-\sum_{n=1}^{\infty}\left(pe^{-\lambda t}\right)^n\right)
\\&=\frac{1-p}{p}\left(\frac1{1-p}-1-\frac{1}{1-pe^{-\lambda t}}+1\right)\end{align*} by applying twice the formula of the geometric series. For the first one to converge, we require $0\le p<1$ and for the second one, we require $|pe^{-\lambda t}|<1$ which already follows from $0\le p<1$ and $t>0$. After some rearranging $$P(X\le t)=\frac{1-e^{-\lambda t}}{1-pe^{-\lambda t}}$$ for $t>0$. 

And, as you said, to find the marginal of $N$, you just have to let $t\to \infty$ which will give you that $N$ has the geometric distribution (starting from $1$) on $\mathbb N_{\ge 1}$.
A: $$P\{N=n,X>t\}=P\{N+n\} -P\{N=n,X\leq t\}$$ $$=(1-p)p^{n-1} -(1-p)p^{n-1}(1-e^{-n\lambda t})=(1-p)p^{n-1}e^{-n\lambda t}.$$ Now sum over $n$ to find $P\{X>t\}$.
