What is $\zeta(i \infty )$ if $ \zeta(\infty)=1 $ and what is its geometric interpretation?

Thanks to user reuns for the following information and insight.

From the functional equation, it is enough to know that $1/|\zeta(1+it)|=o(t^{1/2})$. Indeed, assume this claim for the moment. Then $$\pi e^{-\pi t}\sim\pi/|\sin(\pi it)|=|\Gamma(it)\Gamma(1-it)|.$$ Now, $\Gamma(1-it)=-it\Gamma(-it)$. Now $$|\Gamma(1-it)|=|it\Gamma(-it)|=|t\Gamma(it)|.$$ Combining these results and simplifying, we have $$\frac{\pi^{1/2} e^{-\pi t/2}}{t^{1/2}}\sim |\Gamma(it)|.$$ Putting this in the functional equation yields $$|\zeta(it)|\sim t^{1/2}|\zeta(1+it)|.$$ We can now see the limit cannot exist since $t^{1/2}|\zeta(1+it)|\to\infty$ by the condition $1/|\zeta(1+it)|=o(t^{1/2})$.

The fact that $1/|\zeta(1+it)|=o(t^{1/2})$ follows from theorems $8.9$ (A) and (B) of Titchmarsh's Theory of the Riemann Zeta Function. Here, it is proven that $$\limsup_{t\to\infty}\frac{|\zeta(1+it)|}{\log\log t}\geq e^\gamma\quad\text{and}\quad\liminf_{t\to\infty}\frac{1/|\zeta(1+it)|}{\log\log t}\geq\frac{6}{\pi^2}e^\gamma.$$ In other words, for any $T>0$, there exists a $t_0\geq T$ and a $t_1\geq T$ so that $|\zeta(1+it_0)|$ is arbitrarily small and $|\zeta(1+it_1)|$ is arbitrarily large.

That is correct, it is 1 indeed. The definition of essential singularity on Complex Infinity of the Riemann sphere is not that though! To be essential it must have an essential singularity in f(1/z) in 0! Compare that to definition of essential singularity for finite points: there you use 1/f(z). So "an essential singularity in f(1/z) in 0" means that neither limit in 0 for f(1/z) nor 1/(f(1/z)) should exist. Nice, right? And yes, R. zeta function has essential singularity there.