Does a Central Limit Theorem imply a series is $O(\sqrt{N})$? 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?     
 A: The sharp general result in this direction is the classical 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.
A: There is a notion "big O in probability" that may be what you are bumping up against in the literature. The notation (for a sequence $(Y_n)$ of random variables and a sequence $(b_n)$ of positive constants) $Y_n=O_P(b_n)$ means that the sequence $(Y_n/b_n)$ is stochastically bounded, in that for each $\epsilon>0$ there is a cutoff $C>0$ such that $\Bbb P[|Y_n|/b_n>C]<\epsilon$ for all $n$. The CLT indeed implies that your sequence satisfies $X_N=O_P(\sqrt{N})$.
A: As a complement, may be, the following statement citing from "R. I. Serfling, Approximation Theorems of Mathematical Statistics, John Wiley & Sons, 1980." is helpful for your question. 

