Contradicting results on the convergence of a series involving the Gregory's coefficients

Question

Consider the series : $$I(x)=\sum_{m=0}^{\infty}|G_{m+1}|\sum_{j=0}^{m}(-1)^{j}\binom{m}{j}\cos(2\pi (j+1)x)$$ Where $|G_{m+1}|$ are the absolute Gregory's coefficients ($|G_{m+1}|=(-1)^{m}G_{m+1}$). By the binomial theorem, we have : $$\sum_{j=0}^{m}(-1)^{j}\binom{m}{j}\cos(2\pi (j+1)x)=\frac{1}{2}\left(e^{2\pi i x}(1-e^{2\pi i x})^{m}+e^{-2\pi i x}(1-e^{-2\pi i x})^{m}\right)$$ But since : $$\frac{-z}{\log(1-z)}=1-\sum_{m=1}^{\infty}|G_{m}|z^{m}$$ converges in the unit disk, the series $I(x)$ given above converges for $|1-e^{2\pi i x}|<1$. Explicitly, this is given by : $$I(x)=\frac{\sin(2\pi x)}{2\pi x}-\frac{1}{2}\;\;\;\;\;\;\;\;\left(|1-e^{2\pi i x}|<1\right)$$ However, using the Taylor expansion : $$\cos(2\pi x(j+1))=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k)!}(2\pi x)^{2k}(j+1)^{2k}$$ And the globally convergent series for the Riemann zeta function : $$\zeta(s)=\frac{1}{s-1}+\sum_{m=0}^{\infty}|G_{m+1}|\sum_{j=0}^{m}(-1)^{j}\binom{m}{j}(j+1)^{-s}\;\;\;s\in \mathbb{C}/{1}$$ we have : $$\sum_{m=0}^{\infty}|G_{m+1}|\sum_{j=0}^{m}(-1)^{j}\binom{m}{j}\cos(2\pi (j+1)x)$$$$=\sum_{m=0}^{\infty}|G_{m+1}|\sum_{j=0}^{m}(-1)^{j}\binom{m}{j}\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k)!}(2\pi x)^{2k}(j+1)^{2k}$$ $$=\frac{1}{2}+\sum_{k=1}^{\infty}\frac{(-1)^{k}}{(2k+1)!}(2\pi x)^{2k}=\frac{\sin(2\pi x)}{2\pi x}-\frac{1}{2}\;\;\;\;\;\left(x\in \mathbb{R}\right)$$ where we used : $$\zeta(-2k)=0\;\;\;\;(k=1,2,3...)\;\;\;\;\;\;\zeta(0)=-\frac{1}{2}$$ The two approaches give different regions of convergence ! Where did i go wrong ?

Answer 1

A triple series

$$\sum_{m=0}^{\infty}|G_{m+1}|\sum_{j=0}^{m}(-1)^{j}\binom{m}{j}\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k)!}(2\pi x)^{2k}(j+1)^{2k}$$

can fail to converge absolutely. For instance, for $x=1/2$, when $$\sum_{j=0}^{m}(-1)^{j}\binom{m}{j}\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k)!}(2\pi x)^{2k}(j+1)^{2k}=$$ $$\sum_{j=0}^{m}(-1)^{j}\binom{m}{j}\cos (2\pi(j+1)x)=$$ $$\sum_{j=0}^{m}(-1)^{j}\binom{m}{j}(-1)^{j+1}= -2^m.$$ So we may be not allowed to change the order of its summation. Thus a convergence of the interchanged series can fail to imply a convergence of the initial series.