The problem is:
Prove that $(\forall g: \mathbb{N} \rightarrow \mathbb{R}^+)(\exists f: \mathbb{N} \rightarrow \mathbb{R}^+)(\forall c, B \in \mathbb{R}^+)(\exists n \in \mathbb{N})(n \geq B \wedge f(n) > cg(n))$
My question is that when you assume $g(n)$ can you let $f(n) = cg(n) + 1, c \in R^+$ even though the $\forall c \in R^+$ comes after the $\exists f$. Can you assume that the two $c$'s will be the same?
You want a function $f$ that grows much faster than $g$. Hint: try $f$ of the form $f_1(n) + f_2(g(n))$, where $f_1(n)$ by itself will work if $g(n)$ is bounded, and $f_2(g(n))$ would work if it is unbounded.