The sum of power-law-distributed random variables. Let $X_i$ is power-law-distributed random variable $f(x)=C_0x^{-k}$ where  $1<k_i\le3$. What is the exponent $k$ of the variable 
$$
X=\sum_{i=1}^N X_i \  ?
$$
My doubt come from the fact that $X$ as a sum of i.i.d has to tend to a $\alpha$-stable distribution.
The generic exponent $\alpha$ of a generic $\alpha$-stable distribution can lay only in the range $(0,2]$, that imply $1<k\le 3$. But if we try to use the rule of Fourier transform for the sum of i.i.d. random variables (namely the convolution is the product of the Fourier transforms) as power law we can get an arbitrary big exponent $k$ (isn't it?). So at some point my reasoning is wrong. I guess that the mistake is in the convolution of the power law distributions.

Assume for simplicity that $\bar c_i/x^{k_i-1}\leqslant\mathrm P(X_i\geqslant x)\leqslant c_i/x^{k_i-1}$ when $x\to\infty$ and that each random variable $X_i$ is almost surely nonnegative. Then, for every $1\leqslant j\leqslant N$,
$$
[X_j\geqslant x]\subseteq[X\geqslant x]\subseteq\bigcup_{i=1}^N[X_i\geqslant x/N].
$$
This implies that
$$
\max\limits_{i=1}^N\mathrm P(X_i\geqslant x)\leqslant\mathrm P(X\geqslant x)\leqslant\sum_{i=1}^N\mathrm P(X_i\geqslant x/N),
$$
hence $X$ has exponent $k=\min\limits_{i=1}^Nk_i$ in the sense that, when $x\to+\infty$,
$$
\bar C_N/x^{k-1}\leqslant\mathrm P(X\geqslant x)\leqslant C_N/x^{k-1}.
$$
The result you mention about $\alpha$-stable distribution concerns the regime where $N\to\infty$ and one rescales $X$, hence the constants $C_N$ and $\bar C_N$ come into play and modify the exponent $k$.
