# $\sum\limits_{n=1}^\infty\frac{a_n}{n}$ is convergent implies $\sum\limits_{n=1}^\infty\frac{a_n}{n^{1+it}}$ is convergent

We know that if $$a_n\geq0$$, then obviously, $$\sum\limits_{n=1}^\infty\dfrac{a_n}{n}$$ converges implies $$\sum\limits_{n=1}^\infty\dfrac{a_n}{n^{1+it}}$$ converges for $$t\in\mathbb R$$. But if $$a_n$$ is not real, we can construct a Dirichlet series $$D(s)=\sum\limits_{n=1}^\infty\dfrac{a_n}{n^s}$$ which is convergent at $$s=1$$, but not convergent at $$1+it$$ for some $$t\neq0$$. In fact, let $$p$$ denote prime number. We know that the prime zeta function $$P(s)=\sum_{p\,\in\mathrm{\,primes}} \frac{1}{p^s}=\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{5^s}+\frac{1}{7^s}+\frac{1}{11^s}+\cdots$$ which is convergent for all $$1+it$$ with $$t\neq0$$. Let $$a_n=\begin{cases} 1/p^i& n=p\,\, \text{is prime}\\ 0& \text{otherwise} \end{cases}$$ We have that $$\sum\limits_{n=1}^\infty\frac{a_n}{n^s}$$ is convergent at $$s=1$$(even convergent for all $$s\neq 1-i$$), but divergent at $$s=1-i$$.

Question Restricts $$a_n$$ to be real but does not require $$a_n$$ to be non-negative. Can the convergence of $$\sum\limits_{n=1}^\infty\dfrac{a_n}{n}$$ lead to the convergence of $$\sum\limits_{n=1}^\infty\dfrac{a_n}{n^{1+it}}$$ for all $$t\in\mathbb R$$?. Or can anyone give a counter example

• Try $a_n = \Re(1/p^i)$ Commented Sep 25, 2022 at 6:59
• Look up Dirichlet eta function Commented Sep 25, 2022 at 8:38
• @TravorLZH I know that the eta function is analytic in the whole complex plane, what I'm asking is a general result.
– HGF
Commented Sep 25, 2022 at 8:43
• Let $a_n=\cos(nt)$ Commented Sep 25, 2022 at 8:44
• @HGF If complex sequence $b_n$ converges then $\Re(b_n)$ converges too Commented Sep 25, 2022 at 8:46