martigale convergence theorems Let $S_n = X_{1}+\cdots + X_{n}$ be a martingale satisfying $E[X_{k}^{2}]\leq k<\infty$, for all $k$. Show that $S_{n}$ obeys the weak law of large numbers:
$$P\left(\left|\dfrac{S_{n}}{n}\right|>\epsilon\right)\rightarrow 0,$$ as $n\rightarrow \infty,$ for any positive $\epsilon$.
Note: I used the inequality Markov and got catching up $\dfrac{1}{2\epsilon^{2}}$
$\displaystyle{E[S_{n}^{2}]=\sum_{i=1}^{n}E[X_{i}^{2}]+\sum_{0\leq i<j\leq n}^{n}E[X_{i}X_{j}]}$, but as $S_n$ is martingale we have that $E[X_{i}X_{j}]=0$
Can anyone help me?
 A: Here's some ideas, which are too long for a comment. I don't think it works out completely, but maybe others can patch it up. Since $S_n$ is a martingale, you have $\mathbb{E}S_n = \mathbb{E}S_1 = \mathbb{E}X_1$ for all $n\geq 1$. Applying this in the particular case $n=2$, we get $\mathbb{E}X_1+\mathbb{E}X_2=\mathbb{E}S_2=\mathbb{E}X_1$, and so $\mathbb{E}X_2=0$. Similarly, $\mathbb{E}X_n=0$ for all $n\geq 2$. We don't know if $\mathbb{E}X_1=0$ or not.
The probability you consider is equivalent to $P(|S_n|>n\epsilon)$. We have $|S_n| = |S_n-\mathbb{E}S_n+\mathbb{E}S_n|\leq |S_n-\mathbb{E}S_n|+|\mathbb{E}S_n|=|S_n-\mathbb{E}S_n|+|\mathbb{E}X_1|$.
Thus $P(|S_n|> n\epsilon)\leq P(|S_n-\mathbb{E}S_n|>n\epsilon-|\mathbb{E}X_1|)\leq P(|S_n-\mathbb{E}S_n|\geq n\epsilon-|\mathbb{E}X_1|)$. The right hand side of the last probability will be positive for $n$ sufficiently large.
You write yourself that
$$\text{Var}S_n = \mathbb{E}[S_n^2]-\mathbb{E}[S_n]^2 = \sum_{k=1}^n\mathbb{E}[X_k^2] - \mathbb{E}[X_1]^2.$$
Using the bound $\mathbb{E}[X_k^2]\leq k<\infty$ for all $k\geq 1$, I can bound this variance by
$$\text{Var}[S_n] \leq \sum_{k=1}^n k - \mathbb{E}[X_1]^2=\frac{n(n+1)}{2}-\mathbb{E}[X_1]^2.$$
Now I would like to apply Chebyshev's inequality to the probability $P(|S_n-\mathbb{E}S_n|\geq n\epsilon-|\mathbb{E}X_1|)$, but it doesn't look like it gives me a sharp enough bound.
