Let $X_1, X_2, \ldots $ independent where $X_k$ is exponentially distributed with parameter $1/k!$ i.e. $X_k$ has density
$$f_k(x) = \frac{1}{k!} e^{-\frac{1}{k!} x} 1_{[0,\infty)}$$
For $S_n := \sum_{k = 1}^n X_k$ we want to show that
$$\frac{S_n}{n!} \rightarrow X $$
in law, where $X$ is exponentially distributed with parameter $1$.
There is the hint to consider the distribution of $\frac{X_n}{n!}$.
My first attempt was to use Levy's continuity theorem. The characteristic function of an exponential distribution with parameter $\alpha$ is $\varphi_X(t) = \frac{\alpha}{\alpha – it}$. Furthermorem, for $c \in \mathbb{R}$ we also have $\varphi_{cX}(t) = \frac{\alpha}{\alpha – itc} = \frac{\alpha/c}{\alpha/c – it}$. Since all the variables were independent we have
$$ \varphi_{\frac{S_n}{n!}}(t) = \prod_{k = 1}^n \varphi_{\frac{X_k}{n!}} = \prod_{k = 1}^n \frac{\frac{n!}{k!}}{\frac{n!}{k!} – it} = \prod_{k = 1}^n \frac{n!}{n! – k!it}$$
but now I am struggling to show that indeed $\forall t \in \mathbb{R}: \varphi_{\frac{S_n}{n!}}(t) \rightarrow \frac{1}{1- it}$
Best Answer
For $|t|<n$ (say) we have $\varphi_{S_n/n!}(t)=(1-it)^{-1}(1-it/n)^{-1}\exp f_n(t)$, where $$|f_n(t)|=\left|\sum_{k=1}^{n-2}\log(1-itk!/n!)\right|\leqslant(n-2)\mathcal{O}\left(\frac{|t|}{n(n-1)}\right)\underset{n\to\infty}{\longrightarrow}0.$$ Thus, $\varphi_{S_n/n!}(t)\underset{n\to\infty}{\longrightarrow}(1-it)^{-1}$ (even uniformly if $t$ is bounded).