# Infinite series with only two zeros at $\Re(s)=\frac12$. Why is that the case?

I was experimenting with the following series:

$$\displaystyle k(s)=\sum_{n=1}^\infty \frac{1}{n^{\ln (\frac{n}{s})}}$$

I believe this is an entire function without any poles in $\mathbb{C}$ (note that $s \rightarrow 0 =1$). It also doesn't vanish anywhere, however for $k(s)+k(1-s)$ I found these only two zeros: $s=\frac12 \pm 3.421... i$ and for $k(s)-k(1-s)$ only this pair $s=\frac12 \pm 21.306... i$ and obviously $s=\frac12$.

Can it be explained why these pairs of complex zeros must have $\Re(s)=\frac12$?

It also works when using primes instead of integers. This also induces similar pairs of zeros at $\Re(s)=\frac12$. This outcome actually seems quite independent of the choice of $n$ and I guess the zeros only lying at $\Re(s)=\frac12$ is caused by some symmetry in the series combined with the reflexive nature of $k(s)\pm k(1-s)$.