How would you show that the Riemann Zeta function, $\zeta(s) < 0$ for $s \in (0,1)$? 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 $s \in (0,1)$?
 A: 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
$$
\frac{1}{(2k-1)^s}>\frac{1}{(2k)^s}.
$$
A: 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 $0\lt z\lt1$.
A: 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.
