About absolute convergence of complex series

Question

I know I need to incorporate uniqueness theorem of Dirichlet series to get some kind of contradiction, but don't know how to proceed?

Answer 1

In analogy, for a polynomial $p(z)=a_mz^m+...+a_1z+a_0$ and some number $R>0$, if $|z|>R$, then $1\le R^{-1}|z|$ so that $$ |a_{m-1}z^{m-1}+...+a_1z+a_0|\le \left(|a_{m-1}|+|a_{m-2}|R^{-1}+...+|a_0|R^{m-1}\right)\cdot |z|^{m-1} = A_R\cdot |z|^{m-1}. $$ Then by the triangle inequality $$ |p(z)|\ge|z|^{m-1}\left(|a_m||z|-A_R\right). $$ Thus if additional to $|z|>R$ also $$ |z|>\frac{A_R}{|a_m|}, $$ then $z$ can not be a root of $p$. Or in other words, any root of $p$ satisfies $$ |z|\le\max\left(R,\frac1{|a_m|}\left(|a_{m-1}|+|a_{m-2}|R^{-1}+...+|a_0|R^{m-1}\right)\right). $$

---

To prove your claim, you have to follow similar steps. If $Re(s)>c$, then for $n>m$ $$|n^{s}|=n^{Re(s)}= n^{c}n^{Re(s)-c}\ge n^{c}(m+1)^{Re(s)-c},$$ which implies for the tail of the considered series the bound $$ \left|\sum_{n>m}a_nn^s\right|\le (m+1)^{c-Re(s)}\sum_{n>m}|a_n|n^c=A_c\,(m+1)^{-Re(s)} $$ By the triangle inequality we get $$ |f(s)|\ge |a_m|m^{-Re(s)}-A_c\,(m+1)^{-Re(s)}. $$ As $A$ is finite by assumption, the second term can become smaller than the first term. If that is the case, then $|f(s)|>0$, so $s$ can not be a root of $f$. This dominance condition is equivalent to $$ \left(\frac{m+1}{m}\right)^{Re(s)}>\frac{A_c}{|a_m|}. $$ After taking the logarithm and isolating $Re(s)$, the claim results.