$\zeta(0)=-\frac{1}{2}$ How can I proove $\zeta(0)=-\dfrac{1}{2}$ only with the fact $\zeta(s)=\sum_{n=1}^{\infty}n^{-s}$ for $\mathbb{R}\ni s>1$? Does Euler's formula for $\zeta(2n),~\mathbb{N}\ni n>0$ holds also for $n=0$?
 A: You may exploit the analytic continuation given by the identity:
$$ \sum_{n\geq 1}\frac{(-1)^{n+1}}{n^s} = (1-2^{1-s})\,\zeta(s) $$
that allows us to compute $\zeta(s)$ for $s\in(0,1)$ through:
$$ \zeta(s) = \frac{1}{1-2^{1-s}}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^s}=\frac{1}{1-2^{1-s}}\sum_{n\geq 1}\frac{(-1)^{n+1}}{\Gamma(s)}\int_{0}^{+\infty}u^{s-1}e^{-nu}\,du $$
or:
$$ \zeta(s) = \frac{1}{(1-2^{1-s})\,\Gamma(s)}\int_{0}^{+\infty}\frac{u^{s-1}}{1+e^u}\,du. $$
In order to prove $\zeta(0)=-\frac{1}{2}$, it is enough to show that:
$$ \lim_{s\to 0^+}\frac{\int_{0}^{+\infty}\frac{u^{s-1}}{1+e^u}\,du}{\int_{0}^{+\infty}\frac{u^{s-1}}{e^u}\,du}=\frac{1}{2}$$
but that easily follows from the convergence of $\frac{u^{s-1}}{\Gamma(s)e^u}$ to the Dirac delta distribution $\delta(u)$.
A: One can use the following relationship:
$$\zeta(s)=\frac1{1-2^{1-s}}\eta(s)$$
Where $\eta(s)$ is Dirichlet Eta function,
$$\eta(s)=\sum_{n=0}^\infty(-1)^nn^{-s}$$
In your case, this becomes
$$\eta(0)=\sum_{n=0}^\infty(-1)^n=1-1+1-1+\dots$$
Applying an Euler Transform, we get
$$\eta(0)=\frac12-0+0-0+\dots$$
$$=\frac12$$
So,
$$\zeta(0)=\frac1{1-2^{1-0}}\eta(0)$$
$$=-\frac12$$
EDIT
Or you could just use $\zeta(2n)=(-1)^{n+1}\frac{B_{2n}(2\pi)^{2n}}{2(2n)!}$ to get $\zeta(0)=-\frac12$
