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So SLLN holds and mean of the independent random variables $xi$ is 0, I have to prove that the series of $\frac{\xi_n}{n^{1+\alpha}}$ converges for any $\alpha > 0$.

I'm using 3-series theorem but not sure how to use SLLN to prove that the series $\operatorname{Var} \left( \frac{\xi_n}{n^{1+\alpha}} \right)$ converges. I'd greatly appreciate your help.

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Introduce $ \eta_n=\frac1n\sum_{k=1}^n\xi_k\qquad\text{and}\qquad a_n=n\left(\frac1{n^{1+\alpha}}-\frac1{(n+1)^{1+\alpha}}\right). $ Then, $ \sum\limits_{n\geqslant1}\frac{\xi_n}{n^{1+\alpha}}=\sum\limits_{n\geqslant1}a_n\eta_n. $ The proof is complete, using the following facts:

  1. The sequence $(\eta_n)_{n\geqslant1}$ is almost surely bounded since $\eta_n\to0$ almost surely.
  2. The series $\sum\limits_na_n$ converges since $a_n\geqslant0$ for every $n$ and $ \sum\limits_{n\geqslant1}a_n=\sum\limits_{n\geqslant1}\frac1{n^{1+\alpha}}\leqslant1+\int_1^{+\infty}\frac{\mathrm dx}{x^{1+\alpha}}=1+\frac1{\alpha}. $
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    There is a factor $n$ in the numerator of $a_n$ **AND** the sum of the series $\sum\limits_{n\geqslant1}a_n$ is what I wrote. (Did you try my hint?)2011-12-20