Why is this true ? $(1-s)\zeta(s) = \sum_{k=0}^\infty \frac{\Gamma(k+1-s/2) A_k}{\Gamma(1-s/2) k!}$ On the whole complex plane the following formula is valid :
$$(1-s)\zeta(s) =  \sum_{k=0}^\infty \frac{\Gamma(k+1-s/2) A_k}{\Gamma(1-s/2) k!}$$
where $A_k = \sum_{j=0}^k (2j-1) \zeta(2j+2) (k,j)$
and $(k,j)$ is the binomial coefficient.
How to show this ?
 A: First of all note that you lost $(-1)^{j}$ in $A_{k}.$ And my proof works with $(2j+1)$, so I think there is some problems. However, note that $$A_{k}=\sum_{j=0}^{k}\left(-1\right)^{j}\dbinom{j}{k}\left(2j+1\right)\zeta\left(2j+2\right)=\sum_{j=0}^{k}\left(-1\right)^{j}\dbinom{j}{k}\left(2j+1\right)\sum_{n\geq1}\frac{1}{n^{2j+2}}=\sum_{n\geq1}\frac{1}{n^{2}}\sum_{j=0}^{k}\left(-1\right)^{j}\dbinom{j}{k}\left(2j+1\right)\frac{1}{n^{2j}}.$$
 Now rewrite the last term as $$\left(2j+1\right)\frac{1}{n^{2j}}=n\lim_{a\rightarrow1}\frac{d}{da}\left(\frac{a}{n}\right)^{2j+1}$$
 hence $$A_{k}=\lim_{a\rightarrow1}\frac{d}{da}\left(a\sum_{n\geq1}\frac{1}{n^{2}}\sum_{j=0}^{k}\left(-1\right)^{j}\dbinom{j}{k}\left(2j+1\right)\left(\frac{a^{2}}{n^{2}}\right)^{j}\right)$$ and the now we have a Newton expansion. So$$A_{k}=\lim_{a\rightarrow1}\frac{d}{da}\left(a\sum_{n\geq1}\frac{1}{n^{2}}\left(1-\frac{a^{2}}{n^{2}}\right)^{k}\right).$$
 Using $$\sum_{k \geq 0}\frac{\Gamma\left(k+b\right)}{\Gamma\left(b\right)}\frac{x^{k}}{k!}=\left(1-x\right)^{b}$$
 we have $$\frac{1}{s-1}\sum_{k\geq 0}\frac{\Gamma\left(k+1-s/2\right)}{\Gamma\left(1-s/2\right)}\frac{A_{k}}{k!}=\frac{1}{s-1}\lim_{a\rightarrow1}\frac{d}{da}\left(a\sum_{n\geq1}\frac{1}{n^{2}}\sum_{k\geq0}\frac{\Gamma\left(k+1-s/2\right)}{\Gamma\left(1-s/2\right)}\frac{\left(1-\frac{a^{2}}{n^{2}}\right)^{k}}{k!}\right)=\frac{1}{s-1}\lim_{a\rightarrow1}\frac{d}{da}\left(a^{1+s-2}\sum_{n\geq1}\frac{1}{n^{2}}\left(\frac{1}{n^{2}}\right)^{\frac{s}{2}-1}\right)=\sum_{n\geq1}\frac{1}{n^{s}}=\zeta\left(s\right).$$
A: partial answer
Corrected formula
$$
(s-1)\zeta(s) = \sum_{k=0}^\infty \frac{\Gamma(k+1-s/2) A_k}{\Gamma(1-s/2)k!}
\\
\text{where}\quad A_k = \sum_{j=0}^k (-1)^j (2j+1)\zeta(2j+2)\binom{k}{j}
$$
For $s$ an even positive integer, avoid $\Gamma$-function poles by writing
$$
\frac{\Gamma(k+1-s/2)}{\Gamma(1-s/2)} = (1-s/2)(2-s/2)\cdots(k-s/2)
$$
Then the series is finite, and the identity is true for any function, not only $\zeta$.   
That is, for any function $Z$, and any even positive integer $s$,
$$
(s-1)Z(s) = \sum_{k=0}^\infty \frac{\Gamma(k+1-s/2) A_k}{\Gamma(1-s/2)k!}
\\
\text{where}\quad A_k = \sum_{j=0}^k (-1)^j (2j+1)Z(2j+2)\binom{k}{j}
$$
