Consider the series
$$\sum_{m=0}^{\infty}|G_{m+1}|\sum_{k=0}^{m}(-1)^{k}\binom{m+1}{k+1}f(s,k+1+\delta)$$ Where $|G_{m+1}|$ are the absolute Gregory's coefficients, $0<\delta<1$ and : $$f(s,x)=\frac{x^{-s}}{2\pi i}\left[x^{\frac{2\pi i}{\log j}}\Phi\left(x^{\frac{2\pi i}{\log j}},1,1-\frac{\log j}{2\pi i}s\right)-x^{-\frac{2\pi i}{\log j}}\Phi\left(x^{-\frac{2\pi i}{\log j}},1,1+\frac{\log j}{2\pi i}s\right)\right]$$ where $\Phi(\cdot,\cdot)$ is the Lerch transcendent, $s\in \mathbb{C}$, and $j\in \mathbb{Z^{+}}$. I'm seeking a closed form for this series. I tried expanding $f(s,x)$ as : $$f(s,x)=\frac{x^{-s}}{s\log j}+x^{-s}\sum_{l\in \mathbb{Z}}\frac{x^{\frac{2\pi i l}{\log j}}}{2\pi i l-s\log j}\;\;\;\;\;(A)$$ and i know how to evaluate: $$\sum_{m=0}^{\infty}|G_{m+1}|\sum_{k=0}^{m}(-1)^{k}\binom{m+1}{k+1}(k+1+\delta)^{z}\;\;\;\;z\in \mathbb{C}$$ but the resulting series (A) is only conditionally converging, and i can't reverse the order of the summation. Any help is highly appreciated.
EDIT : we have that : $$\sum_{k=0}^{m}(-1)^{k}\binom{m+1}{k+1}e^{-kx}=(1-e^{x})(1-e^{-x})^{m}+e^{x}$$ Using the generating function of Gregory coefficients : $$\frac{y}{\log(1+y)}=1+\sum_{m=1}^{\infty}G_{m}y^{m}\;\;\;\;\;|y|<1$$ we have : $$\sum_{m=0}^{\infty}|G_{m+1}|(1-e^{-x})^{m}=\frac{1}{1-e^{-x}}-\frac{1}{x}$$ Thus : $$\sum_{m=0}^{\infty}|G_{m+1}|\sum_{k=0}^{m}(-1)^{k}\binom{m+1}{k+1}e^{-kx}=\frac{e^{x}-1}{x}$$ Where we used : $$\sum_{m=0}^{\infty}|G_{m+1}|=1$$ Now we have : $$(1+\delta+k)^{-z}=\frac{1}{\Gamma(z)}\int_{0}^{\infty}e^{-(1+\delta+k)x}x^{z-1}dx\;\;\;\;\Re(z)>0$$ Thus : $$\sum_{m=0}^{\infty}|G_{m+1}|\sum_{k=0}^{m}(-1)^{k}\binom{m+1}{k+1}(1+\delta+k)^{-z}=\frac{1}{\Gamma(z)}\int_{0}^{\infty}(e^{-\delta x}-e^{-(1+\delta)x})x^{z-2}dx$$ $$=\frac{1}{\Gamma(z)}\left(\delta^{1-z}\Gamma(z-1)-(1+\delta)^{1-z}\Gamma(z-1)\right)=\frac{1}{z-1}\left(\delta^{1-z}-(1+\delta)^{1-z}\right)$$ But it can be easily verified that the series above converges everywhere. Thus : $$\sum_{m=0}^{\infty}|G_{m+1}|\sum_{k=0}^{m}(-1)^{k}\binom{m+1}{k+1}(1+\delta+k)^{z}=\frac{1}{z+1}\left[(1+\delta)^{1+z}-\delta^{1+z}\right]\;\;\;z\in \mathbb{C}$$