Real zeros of the zeta function How does one show that the negative even integers make up all the real zeros of the zeta function? That is, how does one show that there are no real zeros on the interval [0,1]?
I am aware that you can in fact compute the zeros numerically, but I am interested in a more noncomputational argument. 
 A: Compairing series and integral, we get
$$\zeta(s) = \int_1^\infty \frac{dt}{t^s} + \sum_{n=1}^\infty \left( \frac{1}{n^s} -\int_{n}^{n+1} \frac{dt}{t^s} \right) = \frac{1}{s-1} + \sum_{n=1}^\infty \left( \frac{1}{n^s} -\int_{n}^{n+1} \frac{dt}{t^s} \right).$$
The RHS is holomorphic for $Re(s) >0$, so equality holds for all $s \in (0,\infty)$.
Let $s \in (0,\infty)$. Since $f(t):=1/t^s$ is decreasing, we have
$$0 \leq \sum_{n=1}^\infty \left( f(n) -\int_{n}^{n+1} f(t)dt \right) < f(1) = 1.$$
Hence $\zeta(s) < \frac{1}{s-1} + 1$, which implies that $\zeta(s)<0$ for $s \in (0,1)$.
A: Analytic continuation of the zeta function to $\mathrm{Re}(s) > 0$ gives (see (16)-(20) here)
$$\zeta(s)=\frac1{1-2^{1-s}}\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}$$
The sum is the Dirichlet $\eta$ function, which can be seen is positive for all real $s > 0$:
$$\eta(s)=\underbrace{\frac1{1^s}-\frac1{2^s}}_{>0}+\underbrace{\frac1{3^s}-\frac1{4^s}}_{>0}+\dots+...\underbrace{\frac1{(2k-1)^s}-\frac1{(2k)^s}}_{>0}+\dots$$
Since $1-2^{1-s}$ is negative for all $s\in(0,1)$, we can conclude that $\zeta(s) < 0$, i.e. $\zeta$ has no zeroes in the range $(0,1)$.  
