On a detail in the proof of: $\zeta(0) = \frac{-1}{2}$

Question

I find a lot of topics concerning $\zeta(0) = \frac{-1}{2}$ but most of them use gamma function $\Gamma$. However, I encouter one problem :

Notice that $\zeta(s) < 0$ for any $0 < s < 1$ where $s \in \mathbb{R}$. Then $$\zeta(0) = \frac{-1}{2}.$$

Since $$\zeta(s) = \frac{1}{1 - 2^{1-s}} \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}$$ for $0 < s < 1.$

Proof Set $$S_N = \sum_{n=1}^N \frac{(-1)^{n-1}}{n^s}.$$ Then $S_{2k} > 1 - \frac{1}{2^s}$ for any $k$. By alternating test, the serie $S_n$ converges, so as any of its subsequence. So $S_{2k}$ converges and its limits greater than or equal to $0$. Since $$\frac{1}{1-2^{1-s}} < 0 $$ for any $s \in (0,1)$, then $$\zeta(s) \leq 0$$ for any $s \in (0,1).$

How to get rid of $=$ from $\leq$?

Another thing is $S_{2k}$ is an increasing sequence, since I want to show that $\zeta(0) = \frac{-1}{2}$, I need to show that $$\lim_{s \rightarrow 0^+} \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} = 1.$$

Note: I also have another form of $\zeta(s), s \in (0,1),$ $$\zeta(s) = 1 + \frac{1}{s-1} - s\int_{1}^\infty \frac{t - [t]}{t^{s+1}}dt$$ where $[t]$ is the integer part of $t$. But I think the form I use above is simpler.

Answer 1

For showing $\eta(0) = \frac12$, an useful way is to show that for $Re(s) > 0$, as $n \to \infty$ : $$n^{-s}-(n+a)^{-s} = \int_n^{n+a}s x^{-s-1}dx = a sn^{-s-1}+ \int_n^{n+a}s (x^{-s-1}-n^{-s-1})dx$$ $$=a sn^{-s-1}- \int_n^{n+a}s \int_n^x (s+1)t^{-s-2}dtdx= a sn^{-s-1}+\mathcal{O}(s n^{-s-2})$$ So that $$n^{-s}-(n+1)^{-s}=\frac{n^{-s}-(n+2)^{-s}}{2}+\mathcal{O}(s n^{-s-2})$$

And hence for $Re(s) > 0$ : $$\eta(s) = \sum_{n=0}^\infty (2n-1)^{-s}-(2n)^{-s} = \sum_{n=0}^\infty \frac{(2n-1)^{-s}-(2n+1)^{-s}}{2}+ \mathcal{O}(s(2n-1)^{-s-2})$$ $$=\frac{1}{2}+\mathcal{O}(s)$$ i.e. $\lim_{s \to 0} \eta(s) = \lim_{s \to 0} \frac{1}{2}+\mathcal{O}(s) = \frac{1}{2}$

Answer 2

We may deduce that identity in a cleaner way. By starting with $$ \begin{eqnarray*}\forall s:\text{Re}(s)>0,\qquad \zeta(s)&=&\left(1-\frac{2}{2^s}\right)^{-1}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^s}\\&=&\left(1-\frac{2}{2^s}\right)^{-1}\frac{1}{\Gamma(s)}\int_{0}^{+\infty}\frac{x^{s-1}}{e^x+1}\,dx\\&\stackrel{IBP}{=}&\frac{4^s}{2(2^s-2)\,\Gamma(s+1)}\int_{0}^{+\infty}\frac{x^s\,dx}{\cosh(x)^2}\end{eqnarray*}$$ the last step of integration by parts extends the domain of convergence to $\text{Re}(s)>-1$, so: $$ \lim_{s\to 0}\zeta(s) = -\frac{1}{2}\int_{0}^{+\infty}\frac{dx}{\cosh^2(x)}=\color{red}{-\frac{1}{2}}$$ since $\int\frac{dx}{\cosh^2(x)}=\tanh(x)$.