Generalized (Gnedenko Kolmogorov) limit theorem for distribution with inverse cube law If $X_1,...,X_n$ are i.i.d.:s with finite variance $\sigma^2$ (and zero mean, for simplicity), the central limit theorem tells us that the stochastic variable $S_n$, defined by $$S_n = (X_1+...+X_n)/\sqrt{n},$$ tends to the Gaussian $N(0,\sigma^2)$ in distribution, as $n \to \infty$.
There is a generalized limit theorem, due to Gnedenko and Kolmogorov, which deals with the case when the $X_i$ are i.i.d.:s with undefined variance (i.e. infinite), see e.g. the following wikipedia article on Stable distribution but also the book "Econophysics" by Stanley and Mantegna. Hence let $X_1,...,X_n$ be i.i.d.:s, drawn from a symmetric distribution around the origin, and with power law tails, such that the pdf $f$ satisfies $$f(x) \sim |x|^{-(1+\alpha)},$$
where $\alpha$ is a parameter. Clearly we must have $\alpha > 0$, in order for the distribution to be normalizable. Further, if $\alpha > 2$, the variance is finite, so the ordinary central limit theorem applies. Hence consider the case $\alpha \in (0,2]$. According to the generalized limit theorem, the stochastic variable $S_n$, defined by $$S_n = (X_1+...+X_n)/n^{1/\alpha},$$
tends to a Lévy $\alpha$-stable distribution, with stability parameter $\alpha$ (the skewness and location parameters are both zero due to the symmetry assumption), as $n \to \infty$.
My question concerns the specific case when $\alpha=2$ and whether the theorem actually applies in this case. Indeed, when $\alpha = 2$ the Lévy $\alpha$-stable distribution reduces to a Gaussian. The generalized limit theorem, as stated above, would therefore imply that if $X_1,...,X_n$ are i.i.d.:s, either with finite variance or with inverse cube tails, then the stochastic variable $S_n$, defined by $$S_n = (X_1+...+X_n)/\sqrt{n},$$ tends to the Gaussian $N(0,\sigma^2)$, as $n \to \infty$. Is this correct, i.e. does the conclusion of the central limit theorem remain valid, even when the variance of the $X_i$ is undefined, provided they are drawn from a distribution with inverse cube tails?
 A: Nope, let $S_n=n^{-1/2}\sum_1^n X_i$ where the $X_i$ are iid and $X_1$ is any symmetric random variable (for your specific question you can take $X_1$ to have pdf $f(x)= |x|^{-3}1_{\{|x|>1\}}$). I claim that $S_n$ won't converge in law to a Gaussian unless $X_1$ has a finite second moment.
For a random variable $Y$, let $\phi_{Y}(x)$ denote the characteristic function of $Y$. It's clear that $$\phi_{S_n}(x) = \phi_{X_1}(n^{-1/2}x)^n.$$
Suppose $S_n$ converged in law to some random variable $Z \sim N(0,\sigma^2)$.
Then the characteristic functions converge pointwise, i.e. $\phi_{S_n}(x) \to \phi_Z(x)$ as $n \to \infty$.
Thus $\phi_{X_1}(n^{-1/2})^n = \phi_{S_n}(1) \to \phi_Z(1) = e^{-\sigma^2/2}>0$. Also, since $X_1$ was assumed to be symmetric, it follows that $\phi_{X_1}$ is real valued. Thus $\phi_{X_1}(n^{-1/2}) > 1-C/n$ for some constant $C$.
Thus $\Bbb E[\cos(n^{-1/2}X_1)] = \phi_{X_1}(n^{-1/2})>1-C/n.$ Thus by Fatou's lemma and the fact that $x^2 = \lim_{t \to 0} \frac{1-\cos(tx)}{t^2/2}$, we find that $$\Bbb E[X_1^2] \leq \liminf_{n \to \infty} \Bbb E \bigg[ \frac{1-\cos(n^{-1/2}X_1)}{\frac12 n^{-1}} \bigg] \leq 2 C. $$
Thus $X_1$ has finite second moment.
Edit: Actually the wikipedia article you linked gives a direct proof of the fact that in the case of $|x|^{-3}$ tails, the partial sums normalized by $\sqrt{n \log n}$ converges to a Gaussian distribution. Notice the additional $\sqrt{\log n}$ factor. You will need to scroll down a little bit in that link.
