If $n\cdot Z_n$ has geometric distribution with parameter $\frac \lambda n$, to what random variable does $Z_n$ converge in distribution to? If $\frac \lambda n$ is the parameter of the geometric distributions then
$$P(nZ_n\le x)=\sum_{i=1}^{[x]} \frac \lambda n \left(1-\frac \lambda n \right)^{i-1} $$
then
$$P(Z_n\le x)=P(nZ_n\le nx)=\sum_{i=1}^{[nx]} \frac \lambda n \left(1-\frac \lambda n \right)^{i-1} $$
I'm guessing that converges in distribution to a Poisson distribution with parameter $\lambda$, but I don't know how to get there.
 A: A lot of times, instead of working with the cumulative distribution functions,  it's easier to work with moment generating functions. 
If you haven't seen these before, for a random variable $X$, the moment generating function is $M_X(t) = E\left(e^{tX}\right)$, if this expected value exists for $t$ in some neighborhood of $0$. One can take advantage of the result that if each of a sequence of random variables $X_n$ has a moment generating function $M_{X_n}(t)$, and $M_{X_n}(t)$ converges everywhere to the moment generating function $M_X(t)$ for some $X$ with known distribution, then $X_n$ converges in distribution to $X$. 
The moment generating function of $nZ_n$ is 
$$M_{nZ_n}(t) = \frac{\frac{\lambda}{n}e^t}{1 - \left(1-\frac{\lambda}{n}\right)e^t}$$
The moment generating function of $Z_n$ is thus
$$M_{Z_n}(t) = \frac{\frac{\lambda}{n}e^{\frac{t}{n}}}{1 - \left(1-\frac{\lambda}{n}\right)e^{\frac{t}{n}}}$$
(One can check from the definition I gave above that moment generating functions transform in this way when you scale a random variable.)
Taking the limit of the latter mgf as $n\rightarrow \infty$ gives 
$$M_{Z}(t) = \frac{\lambda}{\lambda-t}$$
Which happens to be the moment generating function of an Exponential Distribution with parameter $\lambda$. 
