When $\frac{X_n}{n}$ converges almost surely to a negative value, then does it follow that $\sup_\limits{n\ge0}X_n<\infty$ almost surely ?
Are there any further constraints needed to ensure that ?
EDIT: For example $X_n=\sum\limits_{k=1}^{n} Y_k$ where $Y_k$ are iid r.v. with distribution
$P(Y_k=1)=p<\frac12$ and $P(Y_k=-1)=1-p>\frac12$
Let $a$ be the negative a.s. limit of $\frac{X_n}{n}$ and $A=\{\omega:\frac{X_n(\omega)}{n}\to a\}$. Then $P(A)=1$.
If there is $\omega\in A$ such that $\sup_{n\geq0}X_n(\omega)=\infty$, then there is a subsequence $X_{n_k}(\omega)$ such that $X_{n_k}(\omega)\geq 0$ for all $k$ and thus $$\limsup_{n\to\infty} \frac{X_{n_k}(\omega)}{n_k}\geq 0$$ and so $$\limsup_{n\to\infty} \frac{X_{n}(\omega)}{n}\geq 0,$$ contradicting to the choice of $\omega$ since $a<0$. Hence $$\{\omega:\frac{X_n(\omega)}{n}\to a\}\subset \{\omega:\sup_{n\geq0} X_n(\omega)<\infty\},$$ and desired result follows.