$\zeta(0)=-\frac{1}{2}$

Question

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 $n=0$?

Answer 1

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 $\zeta(0)=-\frac12$

Answer 2

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 $\delta(u)$.