Suppose $X_i:\Omega \to \mathbb{R}$ are independent. Consider a function $f(n)$ which goes to $+\infty$ as $n\to \infty$, and so $(\sum_{i=1}^n X_i)/f(n)$ is a generalized average.
My understanding is that if $f(n) \to \infty$ as $n \to \infty$, then the event $E_r = \{ \omega \in \Omega \mid (\sum X_n(\omega))/f(n) \to r \in \mathbb{R} \}$ is a tail event, and so its probability is either zero or one.
Consequently, if $X_n/f_n \xrightarrow{a.s.} X$, then $X$ must be almost surely constant.
So something like $(\sum_{i=1}^n (X_i - \mu_i))/\sigma_i\sqrt{n}$, which we often find to converge in distribution to some nice distribution $D$, will usually diverge almost surely. (Unless $D$ itself is like a delta.)
Can someone help me understand why it should diverge almost surely, across a wide variety of choices for $X_i$?
My attempt: Since we are normalizing $Z_i = (X_i - \mu_i)/\sigma_i$ so that it is zero-centered with a prescribed variance, we should think of $\sum_{i=1}^nZ_i/f(n)$ as something like the partial sums of an alternating series. So it is not surprising that the partial sums don't run off to infinity.
On the other hand, suppose that $\sum_{i=1}^n Z_i /f(n)$ did converge to some random variable almost surely. What would that random variable $X$ possibly look like? It would be quite difficult to describe as a function of $\omega \in \Omega$...our variable $\omega$ would basically have to encode all the information $X_1(\omega), X_2(\omega), \dotsc$, and so it would be hard to describe the limiting r.v. $X$ in a simpler way.
Nevertheless, I don't see why there would usually not exist such a variable $X$...that is, why would $\sum_{i=1}^nZ_i/f(n)$ not settle down, in general? I guess it's saying that because $\sum_{i=1}^n|Z_i(\omega)|/f(n)$ does not converge usually, and because it's not a strictly alternating series, there will be subsequences $\left| \sum_{i=k}^{k+N}|Z_i(\omega)|/f(k+N)\right|>\epsilon$ for arbitrarily high $k$.
As indicated in Kevin's comment, the key is to use the Central limit theorem. Fix M>0 and consider $P(S_n>\sqrt{n}M, i.o)$, i.e, the event that $S_n>\sqrt{n}M$ for infinitely many values of $n$ (the i.o. stands for "infinitely often"). By Fatou's lemma, we have $P(S_n>\sqrt{n}M, i.o)\ge \lim\sup_n P(S_n>\sqrt{n}M)=P(N(0,1)>M)>0$, where the second equality follows from the CLT. Applying Kolmogorov's 0-1 law, we have $P(S_n/\sqrt{n}>M,i.o)=1$. Equivalently, $P(\lim\sup_n S_n/\sqrt{n}>M)=1$.
Finally, recalling that $M$ was arbitrary, we may take a sequence $M_k\to\infty$ and use the fact that a countable intersection of full-measure sets has full measure to see that $P(\forall k \lim\sup_n S_n/\sqrt{n}>M_k)=1$. Now the conclusion follows.