Let $X_n$ be a geometric random variable with parameter $p=\lambda/n$. Compute $P(X_n/n>x)$ $x>0$ and show that as n approaches infinity this probability converges to $P(Y>x)$, where Y is an exponential random variable with parameter $\lambda$. This shows that $X_n/n$ is approximately an exponential random variable.
So I know for a geometric r.v. the $P(X>k)=q^k$ and $q=1-p$. So I think that this means for $X_n$, $q=1-\lambda/n$. What is throwing me off is the $X_n/n$ part. I am not sure how this will affect the probability calculation. For $P(Y>x)$ with parameter $\lambda$ I know this equals $1-F(x)$ where $F(x)=1-e^{-\lambda x}$. Now how will this help me be able to show that $X_n$ converges to $P(Y>x)$?