Let $f(n)$ and $F(n)$ be functions from $\mathbb{Z}^{+} \rightarrow \mathbb{R}$ We say that $f(n)$ has normal order $F(n)$ if for each $\epsilon>0$ the set $$ A(\epsilon)=\{n \in \mathbb{Z}^{+} : (1-\epsilon)F(n)<f(n)<(1+\epsilon)F(n) \} $$ has the property that $\underline{d}(A(\epsilon))=1$ Equivalently, if $B(\epsilon)=\mathbb{Z}^{+} \backslash A(\epsilon)$ then $\underline{d}(B(\epsilon))=0$ (Note that $\underline{d}(X)$ denotes the lower asymptotic density of $X$ .

We say that $f$ has average order $F$ if $$ \sum_{j=1}^n f(j) \sim \sum_{j=1}^n F(j) $$