If we define Riemann zeta function as it is usually defined so that $$\zeta (s)=\sum_{n=1}^{\infty}n^{-s}=\sum_{n=1}^{\infty}e^{-s\ln n}$$ then we can rewrite it as:

$$\zeta (s)=\sum_{n=1}^{\infty}\frac {1}{n^x}\cos(y\ln n) - i\sum_{n=1}^{\infty}\frac {1}{n^x} \sin(y\ln n)=w_1(x,y)-iw_2(x,y)$$:

It is known that for $y=0$ both $w_1$ and $w_2$ converge if $x \in (1,+\infty)$ because when $y=0$ we get domain of absolute convergence of $\zeta$.

Now, the question is what about the set of all $y$ such that both $w_1$ and $w_2$ converge for all $x$ in some interval (where the interval is dependent upon $y$ and is pushed beyond $\Re (s) >1$), in other words, the set of all $y$ such that $\zeta (s)$ defined as $\zeta (s)=\sum_{n=1}^{\infty} e^{-s\ln n}$ converges for all $x$ in some interval, so i raise the following question:

Is it true that for every $y\ne 0$ there is $0<\varepsilon<1$ (or we could write $\varepsilon(y)$ because of the dependence of $\varepsilon$ on $y$) such that $w_1$ and $w_2$ both converge for all $x \in (1-\varepsilon (y),+\infty)$?

So the question is about pushing the boundary of absolute convergence to the natural boundary of convergence of the formula that is used almost everywhere as a definition of Riemann zeta for $\Re (s) >1$.