(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.