Currently, I'm trying to understand how the the equation $$\zeta(s, x+y) = \sum_{k=0}^{\infty} \binom{s+k-1}{s-1} (-y)^{k} \zeta(s+k,x) $$ is derived.
I know that that the following derivative, a shift, is relevant: $$ \frac{\partial}{\partial q} \zeta(s,q) = -s \zeta(s+1,q). $$ Following the approach in the link above, one finds \begin{align}\zeta(s,x+y) &= \sum_{k=0}^{\infty} \frac{y^{k}}{k!} \frac{\partial^{k}}{\partial x^{k}} \zeta(s,x) \\ &= \sum_{k=0}^{\infty} \binom{s+k-1}{s-1} (-y)^{k} \zeta(s+k,x). \qquad \qquad(*) \end{align}
In part, I understand equation $(*)$, but not fully. I see that \begin{align} \frac{1}{k!}\frac{\partial^{k}}{\delta x^{k}} \zeta(s,x) &= (-1)^{k}s(s+1)\dots(s+k-1)\zeta(s+k,x) \\ &= \binom{s+k-1}{s-1} \zeta(s+k,x), \end{align} but there is some part of the equation I don't understand:
- Where does the $x+y$ part come from in $\zeta(s,x+y) $ in $(*)$?
Furthermore, I don't see how the equation $$\zeta(\nu+2) = \sum_{k=0}^{\infty} \binom{k+\nu+1}{k} [ \zeta(k+\nu+2)-1 ] \qquad \qquad (**) $$ results from $(*)$. It is mentioned in the wiki article that $y=-1$ is chosen, but which value of $x$ is then picked?
First, let's remember the definition of the Hurwitz zeta function \begin{eqnarray*} \zeta (s,q)= \sum_{n=0}^{\infty} \frac{1}{(n+q)^s}. \end{eqnarray*} We will also need the following useful formula \begin{eqnarray*} \sum_{k=0}^{\infty} \binom{s+k-1}{k} X^k =\frac{1}{(1-X)^s}. \end{eqnarray*} It is actually a total doddle from here ... invert the order of the plums, do the $k$ plum & ... definition of HZ again.