Let $A\subset \mathbb{N}$. Consider the partial $\zeta$ function $$\sum_{a\in A} \frac{1}{a^s}$$ We know this converges for Re$(s)>1$. Under what conditions of $A$, can this be analyically continued and what would it's analytic continuation be?

If $A = d\mathbb{N}$, all the naturals divisible by $d$, then $$\sum_{a\in A}\frac{1}{a^s} = \frac{1}{d^s}\sum_{n=1}^{\infty}\frac{1}{n^s}$$ and so this can be analytically continued. Moreover, we can say that $$\sum_{a\in A}\frac{1}{a^s} - \frac{1}{d^s}\frac{1}{s-1}$$ will be holomorphic on $\mathbb{C}$.

My guess would be that if $$\lim_{n\to\infty}\frac{|A\cup\{1,\dots,n\}|}{n} = \delta>0$$ then $$\sum_{a\in A}\frac{1}{a^s} - \frac{\delta^s}{s-1}$$ is holomorphic on $\mathbb{C}$.

Is this correct? Is there anything that can be said about sets with zero density?


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