Let $V$ be a set of positive integers, and let: $$\varsigma_{V}\left(x\right)\overset{\textrm{def}}{=}\sum_{v\in V}x^{v}$$ Defining the natural density of $V$ by the limit: $$d\left(V\right)\overset{\textrm{def}}{=}\lim_{N\rightarrow\infty}\frac{\left|V\left(N\right)\right|}{N}$$ where $\left|V\left(N\right)\right|$ is the number of elements of $V$ which are $\leq N$, one consequence of the famous Hardy-Littlewood Tauberian theorem is that, if the limit $\lim_{x\uparrow1}\left(1-x\right)\varsigma_{V}\left(x\right)$ exists, then it is necessarily equal to $d\left(V\right)$.
I was wondering if the converse of this is also true; that is, does the existence of the limit defining $d\left(V\right)$ then guarantee that: $$\lim_{x\uparrow1}\left(1-x\right)\varsigma_{V}\left(x\right)\overset{?}{=}d\left(V\right)$$
A proof of this result (or a reference for one) would be much appreciated.
Using some Abel summation, one finds that $$\zeta_V(x)=(1-x)\sum_n{V(n)x^n}=d(V)(1-x)\sum_{(n+1)x^n}+(1-x)\sum_n{d(n)x^n},$$ where $d(n) << n+1$.
Thus $\zeta_V(x)=\frac{d(V)}{1-x}+o((1-x)^{-2})(1-x)$, hence $(1-x)\zeta_V(x)=d(V)+o(1)$.