Consider the following integral of the Riemann zeta function $$ \zeta ( s) \Gamma ( s) =\ \int _{0}^{\infty }\frac{x^{s-1}}{e^{x} -1} dx. $$ Through substitution, this integral becomes the following:$$ \zeta ( s) \Gamma ( s) =\ \int _{0}^{1}\frac{\ ( -\ln( 1-u))^{s-1}}{u} du . $$ Remembering that $$ \ln( 1-x) \ =\ -\left( \ x+\frac{x^{2}}{2} +\frac{x^{3}}{3} +\ldots\right) $$ Therefore solving this integral we obtain the following:
$$ζ(s)Γ(s)=\frac{1}{s-1}+\frac{s-1}{2s} + \sum_{n=1}^\infty\frac{1}{(s+n)}\left(\sum_{k=1}^{n+1} a(n,k)\left(\prod_{j=1}^{k}(s-j)\right)\right) .$$
Using the properties of the gamma function and the zeta reflection equation :
$$ζ(s)=2^{s}π^{s-1}\sin\left(\frac{πs}{2}\right)ζ(1-s)Γ(1-s). $$
We therefore obtain the following:
$$ ζ(1-s)=2^{s}π^{s-1}\cos\left(\frac{πs}{2}\right)(\frac{1}{s-1}+\frac{s-1}{2s} + \sum_{n=1}^\infty\frac{1}{(s+n)}\left(\sum_{k=1}^{n+1} a(n,k)\left(\prod_{j=1}^{k}(s-j)\right)\right))$$
Lets call
$$ f(s) = \sum_{n=1}^\infty\frac{1}{(s+n)}\left(\sum_{k=1}^{n+1} a(n,k)\left(\prod_{j=1}^{k}(s-j)\right)\right).$$
Therefore $$ζ(1-s) =2^{-s+1}π^{-s} \cos\left(\frac{πs}{2}\right)\left(\frac{1}{s-1}+\frac{s-1}{2s} + f(s)\right).$$ Let us call $X(s) = 2^{-s+1}π^{-s} \cos(\frac{πs}{2})$, therefore $ζ(1-s) =X(s)(\frac{1}{s-1}+\frac{s-1}{2s} + f(s))$.
Let us consider the case where $s= a+it$, therefore considering $ζ(1-s)=0$ and substituting $a=\frac{1}{2}$ in the following: $f(s)=-(\frac{1}{s-1}+\frac{s-1}{2s} )$,
$$ f\left(\frac{1}{2}+it\right)=\frac{1}{2}\left(\frac{5-4t^{2}}{1+4t^{2}}\right)+i\left(\frac{2t}{1+4t^{2}}\right) .$$
But if the actual equation for $f(s)$ is the following:
$$ f(s) = \sum_{n=1}^\infty\frac{1}{(s+n)}\left(\sum_{k=1}^{n+1} a(n,k)\left(\prod_{j=1}^{k}(s-j)\right)\right).$$
Therefore my question is how can we prove that this infinite sum above goes to the same limit?
Note 1: this seems to be a polynomial of infinite order therefore this may be violating some theorem I am not aware of?
Note 2: here below are a few terms of $f(s)$, I have also found the general term of how to calculate the $a(n,k)$ values but have not included here. \begin{align*} f(s)=&\; \frac{1}{s+1}\left( \frac{1}{3}(s-1)+\frac{1}{8}(s-1)(s-2)\right)\\ &+ \frac{1}{s+2}\left( \frac{1}{4}(s-1)+ \frac{1}{6}(s-1)(s-2)+ \frac{1}{48}(s-1)(s-2)(s-3)\right)\\ &+ \frac{1}{s+3}\left( \frac{1}{5}(s-1)+ \frac{13}{72}(s-1)(s-2)+ \frac{1}{24}(s-1)(s-2)(s-3)\right. \\ & +\left. \frac{1}{384}(s-1)(s-2)(s-3)(s-4)\right) + \ldots \,. \end{align*}