Positive logarithmic Dirichlet series force boundary nonvanishing

Suppose first that $F(1+i{t}_0)=0$ for some real $t_0\mathrm{\ne }0$. For $\sigma >1$, absolute convergence gives $\log \mathrm{\mid }F(\sigma +iu)\mathrm{\mid }={\sum }_{n\ge 2}{b}_n{n}^{-\sigma }\cos (u\log n)$, so $\log \text{\hspace{0.17em}⁣}(\mathrm{\mid }F(\sigma ){\mathrm{\mid }}^3\mathrm{\mid }F(\sigma +i{t}_0){\mathrm{\mid }}^4\mathrm{\mid }F(\sigma +2i{t}_0)\mathrm{\mid })={\sum }_{n\ge 2}{b}_n{n}^{-\sigma }(3+4\cos {\theta }_n+\cos (2{\theta }_n))$ with ${\theta }_n=t_0\log n$. Since $3+4\cos \theta +\cos (2\theta )=2(1+\cos \theta {)}^2\ge 0$, the product on the left is at least $1$.

Because $F$ is meromorphic with at most a simple pole at $1$, the factor $F(\sigma )$ grows like $O((\sigma -1{)}^{-1})$ as $\sigma \downarrow 1$, while $F(\sigma +2i{t}_0)$ stays bounded and the zero at $1+i{t}_0$ forces $F(\sigma +i{t}_0)=O(\sigma -1)$. Therefore the product from step 1.1 is $O(\sigma -1)\to 0$, contradicting the lower bound $1$. The Landau statement [L1] concerns singularity of the logarithmic series at its own abscissa, so it does not by itself exclude a zero of $F$ at $1$. Instead, if $F(1)=0$, then $F$ is holomorphic at $1$ and $F(\sigma )\to 0$ as $\sigma \downarrow 1$, whereas the assumed logarithmic identity at $t=0$ gives $\log \mathrm{\mid }F(\sigma )\mathrm{\mid }={\sum }_{n\ge 2}{b}_n{n}^{-\sigma }\ge 0$ and hence $\mathrm{\mid }F(\sigma )\mathrm{\mid }\ge 1$ for every $\sigma >1$. This is another contradiction. Thus no zero occurs on the line $\mathrm{Re}s=1$.