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.



What is the liminf of a sum of i.i.d. random variables with heavy tails?
$\newcommand{\de}{\delta}\newcommand{\ep}{\varepsilon}$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](https://encyclopediaofmath.org/wiki/Attraction_domain_of_a_stable_distribution) and [Stable distribution](https://encyclopediaofmath.org/wiki/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\*}


