I am interested in developing an accurate estimate for the following function: $$ F(A) = \frac{1}{2}+ \frac{1}{\pi}\sum_{t=1}^{\infty}\frac{\sin(A t)}{t}; $$ here $A=4\pi (3/2)^n$, $n\in\mathbb{N}$. I am aware that $F(A)\ge 0$; I am looking for a tight lower bound that is positive for $A$ (or $n$). The sum is convergent by Dirichlet's Test; it is just a matter of obtaining a careful lower bound.
This is part of a larger calculation: I want to show that eventually $E(A)\cdot F(A)>4$, where $E(A)$ is positive and exponential. The case where the series is positive is easy since $E(A)\to\infty$ and if the series is non-negative, $F(A)>1/2$. I'm a bit stuck on the case where the sum is negative. Using Maclaurin series and trig identities, we have $$ F(A)=\frac{1}{2} + \frac{i (\log(1 - e^{i A}) - \log(1-e^{-i A}))}{2\pi} $$ $$=\frac{1}{2}+\frac{1}{\pi}\arctan(\cot(\pi A))$$ $$=1-\{A\}$$I have tried to use the periodicity of the numerator, since $A=2\pi r, r=R/T\in \mathbb{Q}$ to write $F$ as a double sum. Let $S_v=\sum_{t=1}^T \frac{\sin(A t)}{t+v\cdot T}$: then we have
\begin{align}F(A) = \frac{1}{2}+\frac{1}{\pi} \left (\sum_{t=1}^T\frac{\sin(A t)}{t}+\sum_{t=T+1}^{2T}\frac{\sin(A t)}{t}+\sum_{t=2T+1}^{3T}\frac{\sin(A t)}{t}+\cdots \right )\\ = \frac{1}{2} +\frac{1}{\pi} \left (\sum_{t=1}^T\frac{\sin(A t)}{t}+\sum_{t=1}^{T}\frac{\sin(A t)}{t+T}+\sum_{t=1}^{T}\frac{\sin(A t)}{t+2T}+\cdots \right )\\ =\frac{1}{2}+\frac{1}{\pi}\sum_{v=0}^{\infty} \left (\sum_{t=1}^{T}\frac{\sin(A t)}{t+v\cdot T} \right ) = \frac{1}{2}+\frac{1}{\pi}\sum_{v=0}^{\infty}S_v. \end{align} I believe the $S_v$ alternate in sign and have decreasing magnitudes. One approach might be developing a careful bound for $S_0$ and showing $|S_0|>>|S_1|>>|S_2|>>\cdots$ to provide a lower bound. Any help is appreciated.
Since $$\dfrac{\sin Am}m = -\Re\left(i\,\dfrac{e^{iAm}}m\right) = -\Re\dfrac d{dA} \dfrac{e^{iAm}}{m^2},$$ then, applying series representation of the dilogarithm function in the form of $$S(A)=\sum\limits_{m=1}^\infty \dfrac{e^{iAm}}{m^2}=\operatorname{Li_2}(e^{iA}),$$ one can get $$-\Re S'(A)= -\Im \ln(1-e^{iA}) = -\arg(1-e^{iA}) = -\arg\left(2\cos^2\dfrac A2\left(1-i\tan\dfrac A2\right)\right), $$ $$-\Re S'(A) = \dfrac12(\pi-\operatorname{mod}(A,2\pi)),$$
$$F(A) = \dfrac12-\dfrac1\pi \Re S'(A) =1-\operatorname{frac}\left(\dfrac A{2\pi}\right),$$
where $\operatorname{frac} x = x-\lfloor x\rfloor.$
Therefore, a lower bound of $\;F(A)\;$ is $\;\mathbf 0\;$ at $\;A\to 2k\pi-0,\;$ where $\;k\in\mathbb N.\;$