Zeros of alternating zeta function

Question

Is there an intuitive explanation for why all the known non-trivial solutions of $$\frac{1}{1^z} - \frac{1}{2^z} + \frac{1}{3^z} - \frac{1}{4^z} + \cdots=0$$ have a real part of $\frac{1}{2}$?

Answer 1

As pointed out in the comments above, your expression is equal to:

$$\frac{1}{1^s} - \frac{1}{2^s} + \frac{1}{3^s} - \frac{1}{4^s} + \cdots = \zeta(s)(1 - 2^{1-s})$$

To illustrate this further we can look at the matrix multiplication that becomes your series above by creating the following two tables:

First table (red and green) called $\zeta(s)$:

$k \mid n : 1$ else $0$.

Second table (red, green and yellow) called $(1 - 2^{1-s})$:

$n=k: 1$ else if $n=2 \cdot k: -2$ else $0$

Taking the matrix product of the two tables above we get:

The first column then has the numerators of your series.

Some Mathematica code to try:

Clear[n, s];
s = 2;
Sum[1/(2*n + 1)^s - 1/(2*n + 2)^s, {n, 0, Infinity}]
Zeta[s]*(1 - 2^(1 - s))

which when evaluated gives identical answers.

So because the alternating series has $\zeta(s)$ as a factor, the alternating series must also have all the non-trivial zeros of the $\zeta(s)$ function as zeros.

Edit 22.4.2015: But note that $(1 - 2^{1-s})$ has zeros with real part not equal to $\frac{1}{2}$