The Dirichlet eta function is defined as follows:
$$ \eta(s) = \left(1-2^{1-s}\right) \zeta(s). $$
If one replaces $s$ by $1$ in the above formula, we get the apparent "absurdity":
$$ \zeta(1)=\frac{\eta(1)}{1-2^{0}}. $$
The left hand side diverges, while the right hand side is a division by $0$.
How can one interpret such a result? Or is this case somehow handled differently in zeta function theory?
We take the limit as $s$ tends to $1$, and use the fact that the Laurent expansion about $s = 1$ of $1 - 2^{1 - s}$ is \[1 - e^{-(s - 1)\log 2} = 1 - 1 + (\log 2) (s - 1) + O((s - 1)^2),\] while for $\zeta(s)$, we have that \[\zeta(s) = \frac{1}{s - 1} + \gamma_0 + O((s - 1)),\] where $\gamma_0$ is the Euler-Mascheroni constant. So \[\lim_{s \to 1} \eta(s) = \lim_{s \to 1} \left((\log 2) (s - 1) + O((s - 1)^2\right) \left(\frac{1}{s - 1} + \gamma_0 + O((s - 1))\right) = \log 2.\]
In the language of complex analysis, $s = 1$ is a removable singularity of $\eta(s)$: while $\eta(s)$ is not initially defined for $s = 1$, the function \[\begin{cases} \eta(s) & \text{if $\Re(s) > 0$, $s \neq 1$} \\\ \log 2 & \text{if $s = 1$} \end{cases}\] is holomorphic on the open half-plane $\Re(s) > 0$ and agrees with $\eta(s)$ when $s \neq 1$. This is called the holomorphic extension of $\eta(s)$.