# Analytic continuation of a Dirichlet series

Suppose we have a Dirichlet series

$$D(s) = \sum_{n=1}^\infty \frac{a(n)}{n^s}$$

which we know is absolutely convergent for $$Re(s)>1$$. Suppose that we prove that $$\lim_{s\to 1^+}D(s) < \infty$$. Does this imply that $$D(s)$$ continues analytically to some half plane to the left of $$s=1$$, that is to the region $$Re(s) > 1-\epsilon$$ for some positive $$\epsilon$$?

Remark: The case where $$a(n)>0$$ for all $$n$$ (or equivalently finitely many $$n$$) is a classical theorem of Landau. I am interested when this is not the case.

Note that $$\frac{\zeta(2s)}{\zeta(s)}=\sum_{n=1}^{\infty}\frac{\lambda(n)}{n^s},$$ for Re$$(s)>1$$ and that $$\frac{\zeta(2s)}{\zeta(s)}=\sum_{n=1}^{\infty}\frac{\lambda(n)}{n^s}=0$$ at $$s=1$$ (this is equivalent to the prime number theorem). So, your question includes the question of whether $$\frac{\zeta(2s)}{\zeta(s)}$$ can be analytic continued to a half-plane Re$$(s)>1-\epsilon$$, That is, Is there $$\epsilon>0$$ such that $$\zeta(s)\neq 0$$ for Re$$(s)>1-\epsilon$$? This is an open problem, about the zeros of $$\zeta(s)$$ in the critical strip $$0<$$Re$$(s)<1$$.