Showing $\sum_{k\ge 0} e^{-an} \frac{(an)^k}{k!}f(\frac{k}{n}) = \Bbb{E}\left(f\left(\frac{X_1+\cdots + X_n}{n}\right)\right)$ Hello everybody i need to show following  equality

$$\sum_{k\ge 0} e^{-an} \frac{(an)^k}{k!}f(\frac{k}{n}) = \Bbb{E}\left(f\left(\dfrac{X_1+\cdots + X_n}{n}\right)\right)$$

Where $(X_i)_i$ are i.i.d. Poisson random variables with parameter $a$ and $f$ is a bounded continuous function.
I've shown that $S_n:=\sum_{i=1}^n X_i$ is a Poisson random variable with parameter $\alpha n$ with the help of the characteristic functions, indeed if $(X_i)_{i=1}^n$ are i.i.d. Poisson random variables with parameter $\lambda_i$ then we have
\begin{align*}
\Phi_{S_n}&:= \Bbb{E}(e^{iS_nt})\\ 
&=\Bbb{E}(e^{it(X_1+\cdots +X_n)}) \\
&= \Bbb{E}(e^{itX_1}\cdots e^{itX_n}) \\
&=\Bbb{E}(e^{itX_1})\cdots \Bbb{E}(e^{itX_n}), \qquad \text{by independence} \\
&=e^{-(\lambda_1+\cdots + \lambda_n)(1-e^{it})}  
\end{align*}
and we have seen in class that the characteristic function uniquely characterizes the probability distribution, so we can conclude that $S_n$ is distributed as Poisson r.v. with parameter $\lambda_1 + \cdots + \lambda_n$.
In our case the parameters are all the same, so $\sum_{i=1}^n$ is a Poisson r.v. with parameter $an$. I can't rewrite the expression $\Bbb{E}\left(f\left(\dfrac{X_1+\cdots + X_n}{n}\right)\right)$ because i always get something different, could you please give me a hand?

If $X$ is a random variable with values in $\mathbb{N}$, then
$$
{\rm E}[f(X)]=\sum_{k=0}^\infty f(k)P(X=k)
$$
for any 'nice' function $f$. This is the law of the unconscious statistician for discrete random variables.
