What is the liminf of a sum of i.i.d. random variables with heavy tails? Let $\{X_i\}_{i=1}^{\infty}$ be i.i.d. random variables such that: i) $X_i > 0$; and ii) $\textrm{Pr}[X_i > x] \sim x^{-\alpha}$ at large $x$ for some $\alpha \in (0, 1)$. Define the quantities
\begin{equation}
S_n \equiv \frac{X_1 + \cdots + X_n}{n^{1/\alpha}}.
\end{equation}
By Kolmogorov's zero-one law (right?), there is some $\mu \in [0, \infty]$ such that $\liminf_{n \rightarrow \infty} S_n = \mu$ almost surely. I just want to know whether $\mu$ is zero, finite, or infinite (and see a proof).
I haven't been able to find an answer to this question anywhere (note that I'm specifically interested in $\alpha < 1$ and that $S_n$ is normalized differently than the usual mean). For a bit of context, I can easily show/find in textbooks that $\limsup_{n \rightarrow \infty} S_n = \infty$. But those textbooks never seem to mention the liminf. The closest result I could find is that when $\alpha = 1$, $\liminf_{n \rightarrow \infty} S_n = \infty$. But this is shown via truncation + SLLN, and I don't see a way to generalize it to $\alpha < 1$.
More generally, one could define
\begin{equation}
S_n^{(\beta)} \equiv \frac{X_1 + \cdots + X_n}{n^{\beta}}.
\end{equation}
Then all the following are not too difficult to prove: i) for any $\beta < 1/\alpha$, $\lim_{n \rightarrow \infty} S_n^{(\beta)} = \infty$; ii) for any $\beta > 1/\alpha$, $\lim_{n \rightarrow \infty} S_n^{(\beta)} = 0$; and iii) for $\beta = 1/\alpha$, $\limsup_{n \rightarrow \infty} S_n^{(\beta)} = \infty$ (all of these holding almost surely). Thus the question of the liminf when $\beta = 1/\alpha$ is the one missing (to me) piece to understanding the scaling of the partial sums, and I'm really surprised that I haven't been able to find it anywhere.
 A: Take the guy who has the nice scaling properties, which is the inverse gaussian for $\alpha=1/2$.  Then $\frac {S_n}{n^\frac 1 \alpha}$ is constant in distribution.  Since sample of this with n greatly separated are almost independent, and since the distribution has some mass about 0, you must get$\frac {S_n}{n^\frac 1 \alpha} < \epsilon$  i.o. regardless of $\epsilon$.  $$$$  This  argument should work provide the distribution is in the domain of attraction of one of those laws, which I think they are.
A: $\newcommand{\de}{\delta}\newcommand{\ep}{\varepsilon}$Here we implement the idea proposed in mike's answer. The proof below works for all $\alpha\in(0,2]$.
Let $a:=1/\alpha$. Let
\begin{equation*}
    Z_n:=X_1+\cdots+X_n,
\end{equation*}
so that
\begin{equation*}
    S_n=Z_n/n^a. 
\end{equation*}
Take any real $\ep>0$.
We have
\begin{equation*}
    S_n\to S \tag{1}\label{1}
\end{equation*}
in distribution as $n\to\infty$ for some nonnegative random variable (r.v.) $S$
such that $P(S\ge\ep)<1$, so that
\begin{equation*}
    q:=\frac{1+P(S\ge\ep)}2<1; \tag{2}\label{2}
\end{equation*}
see e.g. Attraction domain of a stable distribution and Stable distribution.
For each natural $k$, let $C_k\in(0,\infty)$ be such that $P(S\ge C_k)\le1/k^2$.
It follows from \eqref{1} that for some strictly increasing sequence $(n_k)$ of natural numbers and all natural $k$ we have
\begin{equation*}
    n_{k+1}-n_k\to\infty \tag{3}\label{3}
\end{equation*}
as $k\to\infty$,
\begin{equation*}
    P(S_{n_k}>C_k)\le P(S\ge C_k)+1/k^2\le2/k^2,  
\end{equation*}
and
\begin{equation*}
    \ep_k:=\frac{2\ep n_{k+1}^a-C_k n_k^a}{(n_{k+1}-n_k)^a}>\ep. \tag{4}\label{4}
\end{equation*}
Introducing now the event
\begin{equation*}
    A_m:=\{\forall k\ge m\ S_{n_k}\le C_k\},
\end{equation*}
we have
\begin{equation*}
    1-P(A_m)\le\sum_{k\ge m}P(S_{n_k}>C_k)\le\sum_{k\ge m}2/k^2\to0 \tag{5}\label{5}
\end{equation*}
as $m\to\infty$.
Moreover, letting
\begin{equation*}
    T_k:=\frac{Z_{n_{k+1}}-Z_{n_k}}{(n_{k+1}-n_k)^a}, 
\end{equation*}
we have
\begin{equation*}
\begin{aligned}
&   P(A_m,\forall k\ge m\ S_{n_k}>2\ep) \\ 
    &=P(A_m,\forall k\ge m\ Z_{n_k}>2\ep n_k^a) \\ 
    &\le P(\forall k\ge m\ Z_{n_{k+1}}-Z_{n_k}>2\ep n_{k+1}^a-C_k n_k^a) \\ 
    &=P(\forall k\ge m\ T_k>\ep_k) \\ 
    &\le P(\forall k\ge m\ T_k>\ep) \\ 
    &=\prod_{k=m}^\infty P(T_k>\ep), 
\end{aligned}
\tag{6}\label{6}
\end{equation*}
by \eqref{4} and the independence of the $X_i$'s.
By \eqref{3}, $T_k\to S$ as $k\to\infty$ (cf. \eqref{1}), so that, by \eqref{2}, $P(T_k>\ep)\le q<1$ for all large enough $k$.
So, by \eqref{6}, for all natural $m$ we have $P(A_m,\forall k\ge m\ S_{n_k}>2\ep)=0$ and hence for
\begin{equation}
    B_m:=\{\forall k\ge m\ S_{n_k}>2\ep\}
\end{equation}
we have
\begin{equation*}
    P(B_m)\le1-P(A_m)\to0
\end{equation*}
as $m\to\infty$, by \eqref{5}. But $B_1\subseteq B_2\subseteq\cdots$. So, $P(B_m)=0$ for all natural $m$ and hence
\begin{equation*}
P(\exists m\ \forall k\ge m\ S_{n_k}>2\ep)=P(B_1\cup B_2\cup\cdots)=0. 
\end{equation*}
So,
\begin{equation*}
P(\liminf_{n\to\infty}S_n\le2\ep)\ge P(\forall m\ \exists k\ge m\ S_{n_k}\le2\ep)=1, 
\end{equation*}
for any real $\ep>0$.
Thus,
\begin{equation*}
    P(\liminf_{n\to\infty}S_n=0)=1. 
\end{equation*}
