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

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