The answer is probably well-known, but I cannot find anything definite in the literature. 


Suppose we have the usual ingredients of a CLT, i.e. the series


$$X\_N = \sum\_{n=1}^N x\_n $$


where $x\_n$ are i.i.d.'s. The CLT says that $X\_N/ \sqrt{N}$ approaches a normal distribution. 


Some of the literature states $X\_N = O(\sqrt{N})$. That's the interpretation I would like for my purposes! But it seems more like the CLT implies it is $O(\sqrt{N})$ with probability equal to 1. In mathematical physics, we would be un-inclined to make such a distinction. In pure math, what is the rigorous way to state these things? Are there any delicate issues involved? 



Does a Central Limit Theorem imply a series is $O(\sqrt{N})$?
The sharp general result in this direction is the classical [law of the iterated logarithm](https://en.wikipedia.org/wiki/Law_of_the_iterated_logarithm) (LIL). Suppose, after renormalizing if necessary, that the $x\_n$ are iid with zero mean and unit variance. Then the LIL states that
$$\limsup\_{n \to \infty} \frac{X\_N}{\sqrt{N \log \log N}} = \sqrt{2}, \quad \text{a.s.}$$
In your language, that says that with probability 1, $X\_N$ is $O(\sqrt{N \log \log N})$, and that this cannot be improved to $O(\sqrt{N})$. 


To be more careful, it says that for $P$-almost every $\omega$, there is a finite number $C(\omega)$ such that $|X\_N(\omega)| \le C(\omega) \sqrt{N \log \log N}$ for all $N$.


LIL isn't a direct corollary of CLT, and I believe there are settings where either may hold while the other fails. So I don't think it's true that the CLT "implies" the result you desire, but in any case they are both true in your setting.


