I have a question about Abscissa of Convergence of Dirichlet series. The question is ;
"Let $\sigma_{1}$ and $\sigma_{2}$ be real numbers with $\sigma_{1} \leq \sigma_{2} \leq \sigma_{1}+1 .$ Construct an arithmetic function whose Dirichlet series has abscissa of convergence $=\sigma_{c}=\sigma_{1}$ and abscissa of absolute convergence $\sigma_{a}=\sigma_{2}$."
I found a referrence from the book of Apostol , Here is a theorem and its example.
Theorem 11.10 For any Dirichlet series with $\sigma_{c}$ finite we have
$$
0 \leq \sigma_{a}-\sigma_{c} \leq 1
$$
Proof. It suffices to show that if $\sum f(n) n^{-s_{0}}$ converges for some $s_{0}$ then it converges absolutely for all $s$ with $\bar{\sigma}>\sigma_{0}+1 .$ Let $A$ be an upper bound for the numbers $\left|f(n) n^{-\mathrm{s} 0}\right| .$ Then
$$
\left|\frac{f(n)}{n^{s}}\right|=\left|\frac{f(n)}{n^{s_{0}}}\right|\left|\frac{1}{n^{s-s_{0}}}\right| \leq \frac{A}{n^{\sigma-\sigma_{0}}}
$$
so $\sum\left|f(n) n^{-s}\right|$ converges by comparison with $\sum n^{\sigma_{0}-\sigma}$
EXAMPLE : The series
$$
\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{s}}
$$
converges if $\sigma>0,$ but the convergence is absolute only if $\sigma>1$. Therefore in this example $\sigma_{c}=0$ and $\sigma_{a}=1$
I need an Arithmetic function.How can i approach to the question? How can i develop a solution? Thanks for your answers.
Best Answer
With $c\in [0,1]$, $$f(n) = \pm 1,\qquad |\sum_{n\le x} f(n)-x^c|\le 2$$ then $\sum_n f(n) n^{-s}$ converges for $\Re(s) > c$ and $\sum_n |f(n)|n^{-s}$ converges for $\Re(s) > 1$