Referring to this MO question, i managed to do the following :

We denote by $J(k+1,z)$ the sum : $$J(n+1,z)=\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}\frac{\theta(k+1)}{(k+1)^{z}}$$ and by $S(k+1,z)$ the sum : $$S(n+1,z)=\sum_{k=0}^{n}\frac{\theta(k+1)}{(k+1)^{z}}$$ After some algebra, we obtain : $$J(n+1,z)=\sum_{k=0}^{n}(-1)^{k}\binom{n+1}{k+1}S(k+1,z)$$ Thus : $$\Phi(z)=\sum_{n=0}^{\infty}a_{n}\sum_{k=0}^{n}(-1)^{k}\binom{n+1}{k+1}S(k+1,z)$$ Using the notation in the aforementioned question, and a variant of Perron's formula, we have : $$S(k+1,z)=\frac{1}{2\pi i }\int_{c-i\infty}^{c+i\infty}\Theta(s+z)\frac{(k+1+\delta)^{s}}{s}ds$$ Where $\Re(z)>\beta-c,\;\;\;\;c>0,\;\;\;\;0<\delta<1$.

Ignoring issues of convergence (not justified), we have : $$\Phi(z)=\frac{1}{2\pi i }\int_{c-i\infty}^{c+i\infty}\Theta(s+z)\frac{f(s)}{s}ds\;\;\;\;\;(A)$$ Where we define: $$f(s)=\sum_{n=0}^{\infty}a_{n}\sum_{k=0}^{n}(-1)^{k}\binom{n+1}{k+1}(k+1+\delta)^{s}$$ To take a concrete example, consider the globally convergent series for the Riemann zeta function : $$\zeta(z)-\frac{1}{z-1}=\sum_{n=0}^{\infty}|G_{n+1}|\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}(k+1)^{-z}$$ Where $G_{n}$ are the Gregory coefficients. Now, we have (needs some work) : $$f(s)=\sum_{n=0}^{\infty}|G_{n+1}|\sum_{k=0}^{n}(-1)^{k}\binom{n+1}{k+1}(k+1+\delta)^{s}=\frac{1}{s+1}\left((1+\delta)^{s+1}-\delta^{s+1}\right)=\int_{\delta}^{1+\delta}x^{s}dx$$ Thus, according to the formulae above, we have : $$\zeta(z)-\frac{1}{z-1}=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\int_{\delta}^{1+\delta}\zeta(s+z)\frac{x^{s}}{s}dxds\;\;\;\;\;(B)$$ But : $$\frac{1}{2\pi i }\int_{c-i\infty}^{c+i\infty}\zeta(s+z)\frac{x^{s}}{s}ds=\sum^{'}_{n\leq x}n^{-z}$$ Where the dash on the summation symbol indicates that the term $n = x$, if it occurs, is to be halved. Plugging this in the integral above doesn't yield the LHS !! And i must have done something wrong. I am almost certain that the interchange of the integral and summation in (A) is the source of the error. However, we have that : $$\sum^{'}_{n\leq x }n^{-z}=\zeta(z)-\zeta(z,1+\left \lfloor x \right \rfloor)$$ Where $\zeta(\cdot,\cdot)$ is the Hurwitz zeta function. And we have : $$\lim_{\delta\rightarrow 0^{+}}\int_{\delta}^{1+\delta}(\zeta(z)-\zeta(z,1+x))dx=\zeta(z)-\frac{1}{z-1}$$ In other words, inserting the 'smooth part' of $\zeta(z)-\zeta(z,1+\left \lfloor x \right \rfloor)$ in B yields the correct answer ! there is an element of truth to my approach !

**My question** : What have i done wrong ? and how can we rectify the situation ?