Let $\psi(x)=\sum_{k\leq x}\Lambda(k)$ the Second Chebyshev function, and $\epsilon>0$. I would like to ask
Question. Can you prove or disprove that the series $$\sum_{n=1}^\infty\frac{\psi(n)}{e^n}\sin ns$$ converges uniformly on each strip $ \left\{ s\in\mathbb{C}: \left| \Im s \right|\leq 1-\epsilon \right\} $.
Let a $s=x+iy$ in such strip, from $-(1-\epsilon)\leq y\leq 1+\epsilon$, then $y\leq 1+\epsilon$ and $-(1-\epsilon)\leq y$ thus $-y\leq1-\epsilon$ and $y\leq 1+\epsilon$ and following the computations one can write $$ \left| \frac{\psi(n)e^{-n}}{2i} \right|\cdot \left| e^{in(x+iy)}-e^{-in(x+iy)} \right|\leq \frac{\psi(n)e^{-n}}{2} \left( \left| e^{inx} \right| \left| e^{-ny} \right| + \left| e^{-inx} \right| \left| e^{ny}\right| \right) $$ thus $$ \left|\frac{\psi(n)\sin nz}{e^{n}} \right| \leq \frac{\psi(n)}{2e^n}(e^{n(1-\epsilon)}+e^{n(1-\epsilon)})=\frac{\psi(n)}{e^{n\epsilon}}.$$
Claim. And since $\sum_{n=1}^\infty\frac{n}{e^{\epsilon n}}$ is summable for each $\epsilon>0$ , the Weierstrass theorem implies our that the previous series converges uniformly on the cited closed strip.
Notice that I don't write ..since $\sum_{n=1}^\infty\frac{\psi(n)}{e^{\epsilon n}}$ is summable..., since I believe that it is sufficient study those series in my Claim.
I don't know if previous claim is right, and I don't understand/remember why from the summability as geometric series of the negative exponential one has the statement about the uniformly for the first series.
I know these computations, that Prime Number theorem can be written as equivalence with $\psi(n)=n+o(n)$ as $n\to\infty$. That for a real variable $x>0$ one has $\sum_{n=1}^\infty \frac{1}{e^{nx}}=\frac{e^x}{e^x-1}-1$ uniformly convergent on compact subset of $(0,\infty)$ by Weierstrass, thus one can derive term by term to get $$-\frac{e^x}{e^x-1}=\frac{d}{dx}\frac{1}{e^x-1}=\sum_{n=1}^\infty\frac{d}{dx}e^{-nx}=-\sum_{n=1}^\infty ne^{-nx}$$ uniformly convergent on compact subset of $(0,\infty)$ by Weierstrass (but I don't understand what means the summability versus the uniformly convergence of the series).
Many thanks.