Is it known what the next-to-leading order term is in the variance of the central limit distribution for the average of $N$ variables each of which is distributed according to $P(x) \sim 1/x^{1+\alpha}$ for $1 < \alpha$?

So we have $\bar{x} = N^{-1} \sum_{i=1}^N x_i$ and the distribution of $\bar{x}$, $P_N(\bar{x})$, will have a variance $\sigma_N$ which goes to zero as $N\to \infty$ and the first terms are

$\sigma_N \sim 1 / N^{1-1/\alpha}$ for $1 < \alpha < 2$

$\sigma_N \sim ( \log(N) / N )^{1/2}$ for $\alpha = 2$

$\sigma_N \sim 1 / N^{1/2}$ for $2 < \alpha$

What I couldn't find so far is the form of the next-to-leading order terms in particular for $1 < \alpha \leq 2$ (for $2 < \alpha$ I actually found it). I suspect that for $\alpha = 2$ we will have

$$\sigma_N \sim ( \log(N) / N )^{1/2} ( 1 + c / \log(N) + \ldots)$$

but I haven't found this in any of the probability text books so far.

Is it known what the next-to-leading order term is in the variance of the central limit distribution for the average of $N$ variables each of which is distributed according to $P(x) \sim 1/x^{1+\alpha}$ for $1 < \alpha$?

So we have $\bar{x} = N^{-1} \sum_{i=1}^N x_i$ where $x_i$ are independent and distributed according to $P(x)$, then the distribution of $\bar{x}$, $P_N(\bar{x})$, will have a variance $\sigma_N$ which goes to zero as $N\to \infty$ and the first terms are

$\sigma_N \sim 1 / N^{1-1/\alpha}$ for $1 < \alpha < 2$

$\sigma_N \sim ( \log(N) / N )^{1/2}$ for $\alpha = 2$

$\sigma_N \sim 1 / N^{1/2}$ for $2 < \alpha$

What I couldn't find so far is the form of the next-to-leading order terms in particular for $1 < \alpha \leq 2$ (for $2 < \alpha$ I actually found it). I suspect that for $\alpha = 2$ we will have

$$\sigma_N \sim ( \log(N) / N )^{1/2} ( 1 + c / \log(N) + \ldots)$$

but I haven't found this in any of the probability text books so far.