Determining whether a sequence of averages converges almost surely at all

Question

Consider the problem presented in this post: An example: The average of a sequence of random variables does not converge to the mean of each variable in the sequence almost surely, where it is determined that the sequence of averages $(S_n)_{n=2}^\infty, S_n = \frac{1}{n - 1}\sum_{m=2}^nX_m$, $P(X_m = m) = P(X_m = -m) = \frac{1}{2m\log(m)}, P(X_m = 0) = 1 - \frac{1}{m\log(m)}$ does not converge almost surely to $0$. But what do we know about the general behaviour of the said sequence $(S_n)_{=2}^\infty$? How could we determine whether or not the sequence converges almost surely to any constant $c \in \mathbb{R}$?

Answer 1

Suppose $S_n$ converges to a constant $c \ne 0$? Using Chebychev's Inequality: $$ \Pr(|S_n| > \lambda) \le \lambda^{-2} \mathbb E |S_n|^2 .$$ Now $$ \mathbb E|S_n|^2 = \frac1{(n-1)^2} \sum_{m=2}^n \frac{m^2}{m \log(m)} ,$$ which is bounded by $C/\log(n)$ for some constant $C>0$ if $n \ge 2$, because \begin{align} \sum_{m=2}^n \frac{m^2}{m \log(m)} &\le \sum_{m=2}^{\lfloor \sqrt n \rceil} \frac{m}{\log(m)} + \sum_{m=\lceil \sqrt n \rceil}^n \frac{m}{\log(m)} \\ &\le \sum_{m=2}^{\lfloor \sqrt n \rceil} m + \sum_{m=\lceil \sqrt n \rceil}^n \frac{m}{\log(\sqrt n)} \\ &\le \sum_{m=2}^{\lfloor \sqrt n \rceil} \sqrt n + \sum_{m=1}^n \frac{n}{\log(\sqrt n)} \\ & \le n + 2 \frac{n^2}{\log(n)} \\ & \le \frac{n^2}{\log(n)} + 2 \frac{n^2}{\log(n)} \\ &\le 12 \frac{(n-1)^2}{\log(n)} , \end{align} since $\log(n) \le n$, and $n^2 \le 4(n-1)^2$.

So $\Pr(|S_n| > |c|/2) \to 0 $ as $n \to \infty$.