small o(1) notation It's probably a vey silly question, but I'm confused. Does o(1) simply mean $\lim_{n \to \infty} \frac{f(n)}{\epsilon}=0$ for some $n>N$? 
 A: It probably means $$\lim_{n \to \infty} \frac{f(n)}{1} = 0$$
A: Suppose that $f(n)\in o(1)$ . This can be interpreted using to equivalent definitions:


*

*$\forall$ constant $k>0$ $\exists$ $n_{0}$ such that $\forall n \geq n_{0}$  , $n \epsilon \mathbb{N}$, $|f(n)| \leq k$.

*$\forall$ constant $k > 0$ , $\lim_{n \rightarrow \infty} \frac{f(n)}{k} = 0$ , which implies that $\lim_{n \rightarrow \infty} f(n) = 0$.
A: This should be a comment to chazisop's answer; I don't have enough rep to make it.
Chazisop, your quantifiers in 1 are the wrong way round, in fact there are two problems.  Firstly, saying $\forall k>0 : f(n) \leqslant k$ is simply equivalent to saying $f(n) \leqslant 0$.  The right definition for $o(1)$ is that
$$
\forall k > 0\ \exists N\ \forall n \geqslant N :\; |f(n)| \leqslant k.
$$
Note: the $k$-quantifier appears at the start, this is non-negotiable!  Secondly, notice the absolute value signs around $f(n)$.  If you are only thinking of nonnegative functions (e.g. the running time of an algorithm) you can omit them, but not for arbitrary functions.
To the OP, no, that's not what $o(1)$ means.  There are two problems with what you've written: firstly, what is $\epsilon$? Secondly, what are the $n$ and $N$ supposed to be (I don't mean the $n$ in your limit)?  You need to think about this statement more carefully.
The definition of $f(n)$ being $o(1)$ is that $$\lim _{n \to \infty} f(n) = 0.$$  That means that for all $\epsilon>0$ there exists $N_\epsilon$, depending on $\epsilon$, such that for all $n \geqslant N_\epsilon$ we have $|f(n)| \leqslant \epsilon$.  I guess this definition is probably where your $n>N$ comes from.
