Let $Y_n = n X_n$, where $X_n$ is a non-negative martingale with bounded positive mean (no assumptions of independence). By the martingale convergence theorem $X_n \to X$ for some bounded r.v. $X$. Assume $X > 0$ WP $1$ (in order to avoid counter-examples, as I'm only interested in the technique involved here).
It seems obvious that $Y_n \to \infty$ a.s., but how to prove this rigorously? In particular, the problem I am facing is that $\inf X_n$ can be $0$, so I can't think of a way to establish that for an arbitrary $M> 0$, we can find $n$ large enough so that $P\{Y_n > M\} = 1$. I don't think there's a way to use Borel-Cantelli here either, as the sequence is not independent, and there's no way to say that $\sum_n P(A_n) < \infty$ for the sequence of events $\{Y_n \leq M\}$.
This is not true in general. The constant sequence $X_n = 0$ is a non-negative martingale with bounded mean, and has $nX_n = 0$ as well. For a non-trivial example, if we say $(Z_n)$ is a sequence of i.i.d. Bernoulli random variables with $P(Z_n=0)=P(Z_n=2)=\frac 12$, then $X_n := \prod_{m=1}^n Z_m$ is a non-negative martingale but converges to $0$ almost surely and so does $nX_n$.
If we want to ensure that $\mathbb{E}[\lim X_n] > 0$, let $X_n$ be as in the second example and define $M_0$ to be an independent Bernoulli random variable with $P(M_0 = 0)=P(M_0=1)=\frac 12$. The sequence $M_n = M_0 + X_n$ is a martingale, and $(M_n) \rightarrow M$ where $P(M=0)=P(M=1) = \frac 12$, so $\mathbb{E}[M] = \frac 12 > 0$. The sequence $nM_n$ converges to $0$ on the set $\{M_0 = 0\}$, so does not converge to infinity almost surely.
EDIT: The statement that if $(X_n)$ is a sequence of non-negative random variables with $(X_n) \rightarrow X$ a.s. and $X>0$ a.s., then $(nX_n) \rightarrow \infty$ a.s. is true, and needs no assumptions about $(X_n)$ being a martingale. For almost every $\omega$, there exists $N = N(\omega)$ such that $X_n(\omega) > \frac 12 X(\omega) > 0$ for $n \ge N$, so $\lim n X_n(\omega) \ge \lim \frac 12 n X(\omega) = \infty$.