Let $X_1,X_2,\dots$ be independent and identically distributed random variables with $$\mathbb{P}(X_n=1)=\mathbb{P}(X_n=-1)=1/2$$ for all $n\geq1$. Show that the series $$\sum_nX_n/n$$ converges almost surely.
My idea was to define the random process $(S_n)_{n\geq1}$ by $$S_n = \sum_{k=1}^nX_k/k.$$ It is simple enough to show that this is a martingale, and so if I can show that this process is $L^1$-bounded then I can use Doob's almost sure martingale convergence theorem to obtain the result. But I don't know how to show $L^1$-boundedness (or if it even is possible, in which case I need a new idea), so any help would be great!
Best Answer
Using the hint by @user6247850, we show $L^2$-boundedness of the process. It's not too hard to see that a martingale $S_n$ is $L^2$ bounded if and only if its increments satisfy the following summability condition: $$\sum_{n=1}^\infty \mathbb{E}[(S_n - S_{n-1})^2] < \infty$$ This follows from the fact that for a martingale $S_n \in L^2$, orthogonality of increments implies: $$\mathbb{E}(S_n^2) = \mathbb{E}(S_0^2) + \sum_{k=1}^n \mathbb{E}[(S_k - S_{k-1})^2]$$
Applying this to your martingale of interest, we get $\mathbb{E}[(S_n - S_{n-1})^2] = \frac{1}{n^2}$, so that $$\sum_{n=1}^\infty \mathbb{E}[(S_n - S_{n-1})^2] = \frac{\pi^2}{6} < \infty$$
Thus, $(S_n)$ is an $L^2$-bounded martingale. By Doob's martingale convergence theorem, $S_n$ converges almost surely.