What is the power series expansion for Riemann-Zeta at $0$?

Question

What are the first few terms of the Laurent series expansion of $\zeta(0)$? It gets mentioned here but they only show the first term and I am kind of confused on how they got $-1/2$.

Answer 1

A globally convergent series for the Riemann Zeta except $s=1 $ is $$\zeta\left(s\right)=\frac{1}{1-2^{1-s}}\sum_{n\geq0}\frac{1}{2^{n+1}}\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\left(k+1\right)^{-s} $$ hence $$\zeta\left(0\right)=-\sum_{n\geq0}\frac{1}{2^{n+1}}\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}=-\frac{1}{2}-\sum_{n\geq1}\frac{1}{2^{n+1}}\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}=-\frac{1}{2} $$ by binomial theorem.

Answer 2

We can use the integral, which for $x\gt1$, can be seen to be $$ \begin{align} \int_0^\infty\frac{xt^{x-1}}{e^t+1}\mathrm{d}t &=\int_0^\infty\frac{xt^{x-1}}{1+e^{-t}}e^{-t}\;\mathrm{d}t\\ &=x\sum_{k=1}^\infty(-1)^{k-1}\int_0^\infty t^{x-1}e^{-kt}\;\mathrm{d}t\\ &=x\sum_{k=1}^\infty(-1)^{k-1}k^{-x}\int_0^\infty t^{x-1}e^{-t}\;\mathrm{d}t\\[6pt] &=x\eta(x)\Gamma(x)\\[12pt] &=(1-2^{1-x})\zeta(x)\Gamma(x+1)\tag{1} \end{align} $$ Since $(1)$ is an analytic function of $x$, and we can integrate by parts to get $$ \begin{align} \lim_{x\to0^+}\int_0^\infty\frac{xt^{x-1}}{e^t+1}\mathrm{d}t &=\lim_{x\to0^+}\int_0^\infty\frac{t^xe^t}{(e^t+1)^2}\mathrm{d}t\\ &=\int_1^\infty\frac{\mathrm{d}u}{(u+1)^2}\\ &=\frac{1}{2}\tag{2} \end{align} $$ not only do we see that the integral in $(2)$ converges for $x\gt-1$, but also, by comparing with $(1)$, that for $x=0$, we get by analytic continuation that $$ \zeta(0)=-\frac12\tag{3} $$