 Solution

For sufficiently large $n$, $p_n=\lambda /n$ lies in $(0,1]$. The waiting time has the geometric distribution on $1,2,\dots$, so summing its geometric series giveswhenever $(1-\lambda /n)e^{t/n}<1$. Fix $t<\lambda$. The expansions $e^{t/n}=1+t/n+O(n^{-2})$ show that the numerator is $\lambda /n+O(n^{-2})$ and the denominator is $(\lambda -t)/n+O(n^{-2})$. ThereforeThe convergence theorem for moment-generating functions says that pointwise convergence of finite moment-generating functions on a fixed open interval about zero to the moment-generating function of a probability distribution implies convergence in distribution to that distribution. For example take $|t|<\lambda /2$; all large $n$ satisfy the convergence condition throughout that interval. The limit is the moment-generating function from part (a), proving