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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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1 Answer 1

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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While you proved the convergence of a_n, I think you forgot to include n in the numerator? –  Darwin Dec 20 '11 at 6:24
    
@Darwin, No. Hint: pretend for a moment that $\xi_n=1$ for every $n$, hence $\eta_n=1$ for every $n$. Then look at the formula which expresses the $\xi$ series as an $\eta$ series... –  Did Dec 20 '11 at 8:10
    
But a_n series was supposed to have n in the numerator, right? (the vary last line of your proof). This is because it is defined in the sequence a_n above –  Darwin Dec 20 '11 at 13:30
    
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?) –  Did Dec 20 '11 at 16:21

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