I have the following problem:
Let $(M_n)$ be a bounded martingale and $X_n = \sum_{k=1}^{n} 1/k (M_k - M_{k-1})$.
Show that $(X_n)$ is a martingale that converges almost surely and in $\mathcal{L}^2$.
I already did the martingale and almost surely convergence, but I don't know how to show the $\mathcal{L}^2$ part. Any help is appreciated.
Expand out $X_n^2$,
$$X_n^2 = \sum\limits_{k=1}^n\frac{1}{k^2}(M_k - M_{k-1})^2 + \sum\limits_{j<k}\frac{1}{jk}(M_{j}-M_{j-1})(M_k-M_{k-1})$$
Taking expectations yields,
$$ E[X_n^2] = E\left[\sum\limits_{k=1}^n\frac{1}{k^2}(M_k - M_{k-1})^2\right] + E\left[\sum\limits_{j<k}\frac{1}{jk}(M_{j}-M_{j-1})(M_k-M_{k-1})\right] $$
You can show by repeatedly conditioning on $\mathcal{F}_{n-1}, \mathcal{F}_{n-2}$, etc. that the second expectation vanishes since $\{M_n,\mathcal{F}_n\}$ is a martingale. So by the boundedness of $\{M_n\}$ on the first expectation term we get,
$$ \sup\limits_nE[X_n^2] \leq M\sum\limits_{k=1}^\infty\frac{1}{k^2} < \infty $$
So we can apply Doob's $L^2$ martingale convergence theorem to prove $L^2$ convergence.