Suppose that $X_i$ are iid with mean $0$ and variance $1$. Central limit theorem tells us that: $$\frac{1}{\sqrt{n}}\sum X_i \to \mathcal{N}(0,1)$$ in the sense of convergence of distribution functions. My question is about how much the following statement makes sense: $$\sum X_i \to \mathcal{N}(0,n)$$ Formally - this doesn't make sense as a limit of distributions. The sense that I want to talk about this 'convergence' is the following: suppose that we have some fixed, bounded function $f$, then we can look at the limiting behaviour of $\langle f(\sum_{i=0}^n X_i) \rangle$ (i.e. is it constant, polynomially decreasing, exponentially decreasing) as compared to the limiting behavior of $\langle f(x) \rangle, x \sim \mathcal{N}(0,n)$.
My question is: is there any properties we can enforce on the $X_i$ to ensure that the limiting behaviours of $\langle f\rangle$ are the same. To give an example of why this is not true in general, we can take $X_{i}$ to take values in $\{+1,-1\}$, then if $f$ is a bump function with support avoiding the integer values, $\langle f(\sum_{i=0}^n X_i)\rangle = 0$ identically, whereas $\langle f(x) \rangle, x \sim \mathcal{N}(0,n)$ decays exponentially.
To show that there is some $X_i$ where this claim is true, we can take $X_i$ to be normally distributed themselves, and the claim is true. I want to guess that as long as $X_i$ has positive measure support the claim is true since I haven't been able to think of how a counterexample would arise - but I'm not sure how to proceed