# Kolmogorov’s Three-Series Theorem

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.

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

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 $$1$$.