Zeta and gamma function

Question

The trivial zeros of the Riemann zeta function are of the form $-2n$ where $n$ is a positive whole number. The gamma function goes to infinity for every negative whole number. So, I want to know if there is a closed form of the limit $$\lim_{s\to -2n} \zeta(s)\Gamma(s+1)$$ where $n$ is a positive whole number.

Answer 1

The Taylor series of $\zeta(s)$ about $s=-2n$ is $$\zeta(s)=\sum_{k=0}^\infty \frac{(s+2n)^k\zeta^{(k)}(-2n)}{k!}=\zeta'(-2n)(s+2n)+O((s+2n)^2)$$ we also have the Laurent series expansion of $\Gamma(s)$ about $s=-2n$ which is $$\Gamma(s)=\frac1{(2n)!(s+2n)}+O(1)$$ Multiplying these together we can find the above limit as $$\begin{align} \lim_{s\to-2n}\zeta(s)\Gamma(s+1) &=\lim_{s\to-2n}s\zeta(s)\Gamma(s)\\ &=\lim_{s\to-2n}s\left(\zeta'(-2n)(s+2n)+O((s+2n)^2)\right)\left(\frac1{(2n)!(s+2n)}+O(1)\right)\\ &=\lim_{s\to-2n}s\left(\frac{\zeta'(-2n)}{(2n)!}+O(s+2n)\right)\\ &=-\frac{2n\zeta'(-2n)}{(2n)!}\\ &=-\frac{\zeta'(-2n)}{(2n-1)!}\\ \end{align}$$