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I am looking at a problem where the sum of the individual $X_i$ is $S_n=X_1+\dotsm+X_n$. The probability is given as, $P(X_i=i)=P(X_i=-i)=\frac{i^{-\alpha}}{4}$ and $P(X_i=0)=1-\frac{i^{-\alpha}}{2}$.

The task is to find two functions of $\alpha$ such that $(S_{n}- a_n(\alpha))/b_n(\alpha) \implies N(0,1)$ where $\alpha \in (0,1)$.

By CLT $\frac{S_n-n\mu}{\sigma\sqrt{n}} \implies N(0,1)$

So this implies that $a_n(\alpha)=n\mu$ and $b_n(\alpha)=\sigma\sqrt{n}$.

I startet calculating the expected value, which is found as;

$EX_i=0$

Further for the variance;

$Var(X_i)=\sum_{i=1}^{\infty}\frac{i^{2-\alpha}}{2}$

But when I looked at the converging sum it is clear since $\alpha \in (0,1)$ the sum is divergence.

Therefore am I wondering if I have done the wrong approach, and if so how it should be done.

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    $\begingroup$ For $\alpha \le 3$ the variance of $S_n$ is diverging, so be careful. Maybe check the Lindeberg conditions. $\endgroup$ – kjetil b halvorsen Dec 10 '18 at 7:59
  • $\begingroup$ Hint: although the variance of $S_n$ may be expressed as a finite (not infinite!) sum with no simple closed form, it can be closely approximated by an integral whose value asymptotically is $n^{3-\alpha}/(3-\alpha).$ What happens when you use this variance to standardize $S_n$? $\endgroup$ – whuber Dec 10 '18 at 15:42

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