On The Next Bit Test I would like to know what $O(v(n))$ really means in detailed and simple words please.
I found it everywhere in the literature I am reviewing but I cannot find what the intuition of it (especially if it means the Big O Notation, then what it has to do with the Probability here; also what is $v$, it is never defined except in another paper as a constant -but maybe not related directly to v in this formula-).

Many Thanks.
References 1 , 2
But this notation is defined (informally) in the first paper.

The notation $O(\nu(n))$ is used for any
function, $f(n)$, that vanishes faster than the inverse
of any polynomial, that is for every polynomial, $\mathrm{poly}
(n)$, and $n$ large enough, $f(n) \leq 1/\mathrm{poly}(n)$

Therefore, what it means is no probabilistic polynomial time (PPT) algorithm $A$ can guess the next bit at inverse polynomially decreasing error rate.
Given any PPT algorithm $A$ this error probability decays super-polynomially, for example at a rate $\exp(-\log^2 n)$ which goes to zero faster than any polynomial.
Claim: $\exp(-\log^2 n)$ is less than $n^{-c}$ for any constant $c$ for $n$ large enough.
Proof: Look at the reciprocal. $$\exp(\log^2 n) &gt; n^c$$ if and only if
$$\log^2 n &gt; c \log n$$ which will clearly happen as soon as $$\frac{\log^2 n}{\log n}&gt;c$$ i.e., as soon as $\log n&gt;c$.
Note that $\exp(\log^{1+\epsilon} n)$ is superpolynomial for all $\epsilon&gt;0.$
Edit: The superpolynomial convergence is essentially what is referred to as negligibility.