Riemann Zeta Function at $s = 1 + 2 \pi i n / \ln 2$

Question

I am aware that the function defined by $$ \zeta(s) = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots $$ for $Re(s)>1$ can be extended to a function defined for $Re(s)>0$ by writing $$ \zeta(s) = \frac{1}{1-2^{1-s}} \cdot \left( \frac{1}{1^s} - \frac{1}{2^s} + \frac{1}{3^s} - \frac{1}{4^s} + \cdots \right) $$ Using this representation, one sees that at $s=1$, since the sum evaluates to $\ln 2$, the Riemann Zeta Function has a simple pole of residue $1$. However, what happens at $$s= 1 + \frac{2\pi i n}{\ln 2} $$ where $n$ is a non-zero integer? The denominator is zero at these values of $s$, but it is well-known that the Riemann Zeta Function only has the one pole at $s=1$, so the the sum at these values must evaluate to $0$ to eliminate the poles which would otherwise form. How do we know though that that alternating sum has zeroes for these values?

Answer 1

This issue has been adressed by J. Sondow (2003) here: Zeros of the alternating zeta function on the line R(s)=1.