= Solution
For sufficiently large $n$, $p_n=\lambda/n$ lies in $(0,1]$. The waiting time has the <geometric distribution> on $1,2,\ldots$, so summing its <geometric series> gives
$$
M_{Y_n/n}(t)=\sum_{j=1}^\infty p_n(1-p_n)^{j-1}e^{tj/n}
=\boxed{\frac{(\lambda/n)e^{t/n}}{1-(1-\lambda/n)e^{t/n}}},
$$
whenever $(1-\lambda/n)e^{t/n}<1$. 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 $(\lambda-t)/n+O(n^{-2})$. Therefore
$$
M_{Y_n/n}(t)\longrightarrow\frac\lambda{\lambda-t}.
$$
The <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
$$
\boxed{Y_n/n\xrightarrow{d}\operatorname{Exp}(\lambda)}.
$$
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