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Consider the sequence r.v. $X_i's, i \geq 2$, $$P(X_i = i^2) = P(X_i = -i^2) = 1/i^2 \ \text{and} \ P(X_i = (-1)^i) = 1- 2/i^2.$$ Consider $S_n = \sum_{i=2}^{n}X_i$. What is the almost sure limit of $S_n/n$ as $n \rightarrow \infty$?

I have tried to truncate $X_i$ as defining $Y_i = X_i \boldsymbol{1}_{[|X_i| \leq 1]}$. However, not sure how to find the limit and prove $S_n/n$ converges almost surely.

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  • $\begingroup$ Three series Theorem is about convergence of the series $\sum X_i$, not about convergence of the sequence $\frac {S_n} n$. $\endgroup$ – Kavi Rama Murthy Dec 9 '19 at 7:33
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This is not a question on the Three Series Theorem.

Since $\sum P(X_i=i^{2}) <\infty$ and $\sum P(X_i=-i^{2}) <\infty$ Borel Cantelli Lemma tells us that $P(X_i=i^{2} i.o )=0$ and $P(X_i=-i^{2} i.o )=0$ so $X_i=(-1)^{i}$ for all $i$ sufficiently large, with probability $1$. This implies that $(S_n)$ is bounded with probability $1$ so $\frac {S_n} n \to 0$ with probability $1$.

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