For arithmetical periodic function $f$, if $\sum_{r=1}^k f(r)=0$, then $S=\sum_{n=1}^\infty \frac{f(n)}{n^{s}}$ converges

Question

[Introduction to Analytic Number Theory - Tom M. Apostol, chapter 12, question 1(b)]

Let $f(n)$ be an arithmetical function which is periodic mod $k$. If $$\sum_{r=1}^k f(r)=0$$ then prove that the Dirichlet series $$S=\sum_{n=1}^\infty \frac{f(n)}{n^{s}}$$ converges for $\sigma>0$, and that $f$ can be extended to an entire function $F(s)$.

In a previous exercise I proved that $S$ converges absolutely for $\sigma>1$ and $$S=\frac 1{k^s} \sum_{r=1}^k f(r)\zeta\left (s,\frac rk\right )$$ if $\sigma>1$.

I can't figure out how to use the extra assumption of $\sum_{r=1}^k f(r)=0$ to get the behavior in $\sigma>0$.

Also, once the convergence is proved, using the previous exercise, this sum can be expressed as a finite linear combination of Hurwitz zeta functions, and hence is analytic for all $s$ except for $s=1$. How to get rid of this one point?