Have all the zeros of the Riemann Zeta function real part smaller than 1? I think that all the zeros of the Riemann-Zeta function ${\zeta}( z ) = \frac{1}{1-2^{1-z}} \sum_{n = 0}^{\infty} \frac{1}{2^{n+1}} \sum_{k = 0}^{n} (-1)^k \binom{n}{k} (k+1)^{-z}$  have real part smaller than 1, but I don't know how to prove it.
If Re(z) > 1 than 
${\zeta}( z ) = \sum_{n = 1}^{\infty} \left( \frac{1}{n} \right)^z$ 
 A: Proposition. $\zeta(s) \neq 0$ for real bigger than $1$. 
Proof. For real part $> 1$, this follows very quickly from the Euler product,
$$ \zeta(s) = \prod_p \left( 1 - p^{-s}\right)^{-1}.$$
My favorite way of seeing this is to consider
$$\zeta(s) \cdot \prod_p (1 - p^{-s}) = 1.$$
The key here is that for real part of $s$ greater than $1$, we know that $\displaystyle \sum p^{-s}$ converges absolutely, and thus $\displaystyle \prod_p(1 - p^{-s})$ converges absolutely. In particular, it's finite. So $\zeta(s) \neq 0$, since we never have $(0 \cdot \text{finite}) \neq 0$. $\diamondsuit$
It's a bit more annoying for real part of $s$ equal to $1$. The original ways to do this are uninspired, and revolve around a cosine triple angle identity (or something like that). We will do something a bit more interesting, but perhaps only more interesting because slight modifications make it work in much more general contexts.
Proposition. Let $\sigma_a(n) = \displaystyle \sum_{d \mid n} d^a$ be a divisor function. I use $d \mid n$ to mean "$d$ divides $n$," so this is a sum across the divisors of $n$. I also only mean the positive divisors. Then
$$ \sum_{n \geq 1} \frac{\sigma_a(n)\sigma_b(n)}{n^s} = \frac{\zeta(s)\zeta(s-a)\zeta(s-b)\zeta(s-a-b)}{\zeta(2s - a - b)}.$$
Proof. The idea is to find both sides Euler products. We know the right sides Euler product already. The left side has one because $\sigma_a(n)$ is multiplicative.
So the left hand side is given by 
$$\prod_p \left( \sum_{k \geq 0} \frac{\sigma_a(p^k)\sigma_b(p^k)}{p^{sk}} \right).$$
The next step is to realize that $\sigma_a(p^k)$ is a geometric series, so we can just write it down explicitly, $\sigma_a(p^k) = \dfrac{1 - p^{a(k+1)}}{1 - p^{a}}$. The remainder of this proof, I'll write as a sketch. 


*

*Substitute in these geometric series expressions for the sigmas in the summation.

*Multiply everything out. You'll end up with some common factors and four sums in $k$.

*Each of these sums is also geometric. So just write down what they are. Now there is no more $k$ summation.

*Find common denominators and add everything together. If no arithmetic errors were made, many things will cancel.


And we'll end up with
$$\prod_p \left( \frac{1-p^{a + b - 2s}}{(1-p^{-s})(1-p^{a - s})(1- p^{b - s})(1- p^{a + b - s})}\right),$$
which is exactly the Euler product of the right hand side. $\diamondsuit$
Theorem. $\zeta(1 + it) \neq 0$ for real $t$, i.e. $\zeta(s)$ has no zeroes on the line $\Re(s) = 1$.
Proof. Suppose not. So suppose that $\zeta(1+it) = 0$. Then we also know that $\zeta(1 - it) = 0$ by conjugation. Let $a = it$ and $b = -it$, and use the proposition above. Then
$$ Z(s) := \sum_{n \geq 1} \frac{|\sigma_{it}(n)|^2}{n^s} = \frac{\zeta(s)^2\zeta(s-it)\zeta(s+it)}{\zeta(2s)}.$$
Notice that at $s = 1$, the numerator of the right side has two poles from $\zeta(s)^2$ and (at least) two zeroes from $\zeta(s \pm it)$, while the numerator is finite. So it converges. The left side is a series of positive real terms when $s$ is real. We know that the abscissa of convergence on the real axis of the left hand side will be a singularity, and the next candidate singularities are $s \approx \frac{1}{2}$ from the denominator. But in particular, $Z(s)$ is holomorphic for $\Re(s) > \frac{1}{2}$.
We know that $\zeta(s)$ has a pole at $s = 1$. So $\zeta(2s)$ has a pole at $s = \frac{1}{2}$. So if $s = \frac{1}{2} + \delta$ for small $\delta$, the numerator of the RHS expression of $Z(s)$ is bounded, and the denominator tends to $\infty$ as $\delta \to 0$. Hence $Z(s) \to 0$ as $\delta \to 0$.
On the other hand, the LHS expression of $Z(s)$, which we know converges at least for $\Re(s) > \frac{1}{2}$ (and thus for $s = \frac{1}{2} + \delta$ for small positive $\delta$), is at least $1$ everywhere it converges for real $s$ (since the first term is $1$, and all terms are real and positive if $s$ is real). Thus $Z(s) \geq 1$ as $\delta \to 0$.
This is a contradiction, so we have that $\zeta(1 + it) \neq 0$. $\diamondsuit$.
