How would you show that the Riemann Zeta function, $\zeta(s) < 0$ for $s \in (0,1)$?

Question

How would you show that the Riemann Zeta function, $\zeta(s) < 0$ for $s \in (0,1)$?

So far I have that along the critical strip

\begin{align} \zeta(s) &= \left(\frac{2^{s-1}}{2^{s-1}-1}\right)\phi(s)\\ &= \left(\frac{1}{1-2^{1-s}}\right)\phi(s)\\ &= \left(\frac{1}{1-2^{1-s}}\right)\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^s} \end{align}

Where $\phi(s)$ is Euler's alternating zeta function. (which converges for $s > 0$) How would you show that this is always negative when $s \in (0,1)$?

Answer 1

If $s\in(0,1)$ then $\frac{1}{1-2^{1-s}}<0$, while $$\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}}{n^s}$$ is positive due to Leibniz' criterion.

Answer 2

Note that $$ \varphi(s)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^s}=\sum_{k=1}^\infty \left(\frac{1}{(2k-1)^s}-\frac{1}{(2k)^s}\right)>0, $$ as $$ \frac{1}{(2k-1)^s}>\frac{1}{(2k)^s}. $$

Answer 3

Here is an approach using the integral for $\zeta(z)\Gamma(z)$ and a simple analytic continuation.

Waxing complex: For $\mathrm{Re}(z)\gt1$, we have $$ \begin{align} \zeta(z)\Gamma(z) &=\int_0^\infty\frac{t^{z-1}}{e^t-1}\mathrm{d}t\\ &=\frac1{z-1}\int_0^\infty\frac{t}{e^t-1}\mathrm{d}t^{z-1}\\ &=\frac1{z-1}\int_0^\infty t^{z-1}\frac{1+(t-1)e^t}{(e^t-1)^2}\mathrm{d}t\tag{1} \end{align} $$ Since $\frac{\mathrm{d}}{\mathrm{d}t}\left[1+(t-1)e^t\right]=te^t\ge0$ for $t\ge0$, the integrand is positive. The integral in $(1)$ also converges for $\mathrm{Re}(z)\gt0$ and is analytic. Therefore, $(1)$ represents the analytic continuation of $\zeta(z)\Gamma(z)$ to $\mathrm{Re}(z)\gt0$.

Back to the real world: Since $\Gamma(z)\gt0$ for $z\gt0$, $(1)$ says that $\zeta(z)\lt0$ for $0\lt z\lt1$.