# Almsot surely convergence of series

(a) Suppose that $$X_1,X_2,...$$ be independent with $$P(X_n=n-1)=\frac{1}{n}$$, $$P(X_n=-1)=1-\frac{1}{n}$$. Show that there are no constants $${\mu_n}$$ such that $$\frac{s_n}{n}-\mu_n \rightarrow 0$$ a.s.

(b) Suppose that in (a) we replace $$n,\frac{1}{n}$$ throughout by $$n^2,\frac{1}{n^2}$$. show that $$S_n/n \rightarrow -1$$ a.s.

My idea: By Theorem 2.2.7 of Durrett's book, $$x P(|X_i|>x)\rightarrow 0$$ as $$x \rightarrow \infty$$ is a necessary condition for existing $$\mu_n$$ such that $$S_n/n - \mu_n\rightarrow 0$$. But in the Theorem $$X_i$$'s are i.i.d. In this problem $$X_i$$'s don't have same distribution. Is there any way to adopt this theorem?

Durett's theorem cannot be used here. Let us first note that b) is very easy:$$\sum P\{X_n=n^{2}-1\} <\infty$$ so $$P\{X_n=n^{2}-1 \, i.o. \}=0$$ which shows that $$X_n=-1$$ for all $$n$$ sufficiently large with probability $$1$$. Hence $$\frac {S_n} n \to -1$$ almost surely. For part a) we have to find a way of getting rid of $$\mu_n$$'s. Note that $$\frac {S_n} n -\mu_n \to 0$$ almost surely implies that $$\frac {X_n} n -c_n \to 0$$ almost surely where $$c_n=\frac {n-1} n \mu_{n-1} -\mu_n$$. We now get rid of $$c_n$$'s by taking an independent copy $$\{Y_n\}$$ of $$\{X_n\}$$. [ The new sequence is another independent sequence with the same distribution as the original one and independent of the original one]. It follows then that $$\frac {X_n-Y_n} n \to 0$$ almost surely. However $$\{X_n-Y_n\}$$ is an independent sequence such that $$P\{X_n-Y_n=n\}=\frac 1 n (1-\frac 1 n)$$,$$P\{X_n-Y_n=-n\}=\frac 1 n (1-\frac 1 n)$$ and the only other value of $$X_n-Y_n$$ is $$0$$. Since $$\sum P\{X_n-Y_n=n\}=\infty$$ we see that $$X_n-Y_n=n$$ infinitely often with probability $$1$$ contradicting the fact that $$\frac {X_n-Y_n} n \to 0$$ almost surely.