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
