Boundness of Riemann zeta by log function

Question

In the picture above it states that for $Re(s)>1$ we have that lower bound of a is $\leq 0$, and it claims that it is obvious from Dirichlet series of $\log\varsigma(s)=\sum_{n\geq 2}\frac{\Lambda(n)}{n^slog(n)}$. I dont see how it follows from it :/ Can anyone clerify this?

Answer 1

Let $s=\sigma+it$, so that

$$ |\log\zeta(s)|\le\sum_{n\ge2}{\Lambda(n)\over n^\sigma\log n}\le\zeta(\sigma) $$

Since the RHS converges whenever $\sigma>1$, we conclude that $\nu(\sigma)\le0$ for $\sigma>1$.