$\limsup_{k\rightarrow\infty}X_k/k$ for identically distributed random variables $X_k$. Let $\{X_k\}_{k\in\mathbb{N}}$ be a sequence of identically distributed and not independent random variables on the natural numbers. I am interested in conditions that would ensure that almost surely
$$
\limsup_{k\rightarrow\infty}\frac{X_k}{k} < \infty.
$$
One such a condition is $\mathbb{E}(X_k)<\infty$, since then
$$
\sum_{k\in\mathbb{N}}P(X_k\ge k) = \mathbb{E}(X_k)<\infty.
$$
Borel-Cantelli then gives $P(\limsup_{k\rightarrow\infty}\{\omega\in\Omega\mid\ X_k\ge k\}) = 0$, which means that for each $\omega$ there exists $N\in\mathbb{N}$ such that $X_k<k$ for all $k\ge N$. Therefore we get that $\limsup_{k\rightarrow\infty}\frac{X_k}{k}\le 1$.
The random variables I have to deal with are quite complex and thus I am interested in a less strict condition that ensures the finite limit superior. 
Thank you in advance!

EDIT
My thoughts on extended conditions. In the expectation proof and the answer given we look for an $r$ such that $\sum_{k\in\mathbb{N}}P(X_k>rk)<\infty$. I will try to explore whether some random variable replacing $r$ helps, since we dont need the limit superior to be equal for all $\omega$, just finite.
 A: I just saw your comment; note here I'm assuming the $X_k$ are independent. Relevant or not, seems interesting.
The following seems like it must be wrong, can't be that simple. But I don't see where the error is. Suppose $X_k\ge0$. Seems to me that if $\Bbb E[X_k]<\infty$ then $\limsup X_k/k=0$ almost surely, while if $\Bbb E[X_k]=\infty$ then $\limsup X_k/k=\infty$ almost surely.
First, if $\Bbb E[X_k]<\infty$: You've shown that $\limsup X_k/k\le 1$ almost surely. The same applies to $NX_k$ for any positive integer $N$, so that $\limsup X_k/k\le1/N$ almost surely, hence $\limsup X_k/k=0$ almost surely.
On the other hand, note that for $X\ge0$ we have $\Bbb E[X]<\infty$ if and only if $\sum P(X\ge k)<\infty$, regardless of whether $X$ is integer-valued. So if the $X_k\ge 0$ are iid and $\Bbb E[X_k]=\infty $ then Borel-Cantelli shows that $\limsup X_k/k\ge1$ almost surely, and again applying this to $cX_k$ shows that in fact $\limsup X_k/k=\infty$ almost surely.
Edit: Maybe this actually is right! I didn't see how at first, but in fact it's equivalent to the condition that Robert Israel gave in the answer he deleted on grounds of irrelevance. Saying $\sum\ln F(rk)>-\infty$ for some $r>0$ is the same as $\sum \ln F(k)>-\infty$, by monotonicity. Since $F(k)\to1$ from below this is the same as $\sum(1-F(k))<\infty$, which is the same as $\sum P(X_k\ge k)<\infty$.
