Correct me if I phrased the title wrong. In 3Blue1Brown's video about visualizing the zeta function, it can be seen that $\zeta(s)$ (Defined by the infinite series) appears to stop at a line slightly to the right of the line $\Re(s)=\frac{1}{2}$ after the transformation.
This line is actually referenced in the code for the video as $\Re(s)=\gamma$, which was surprising to me. It's odd that the full transformation appears to be roughly symmetrical around that constant specifically too, though it reminds me of limits involving the Zeta function where letting s approach 1 yields $\gamma$. How do you show that the range of $\Re(\zeta(s))$ over $(1,\infty)$ is $(\gamma,\infty)$? Or alternatively, how do you show that the absolute minimum of $\zeta(s)$ over $(1,\infty)$ is $\gamma$? At first, I thought to use the following:
$$\Re(\zeta(a+bi))=\sum_{n=1}^{\infty}\frac{\cos(b\ln(n))}{n^{a}}$$
I think treating it as a multivariable function $f(a,b)$ and finding the minimum might yield $\gamma$? That is hard to do considering that the function is defined as an infinite sum though. Another thought I had was to use this limit: $$\lim_{a\to1+} \sum_{n=1}^{\infty}\frac{\cos(b\ln(n))}{n^{a}};(b\neq0)$$ Visually the left side of the horizontal lines appear to shift over from $1$ to $\gamma$ in the video, which is why I thought this limit may work. However. this doesn't look like any of the definitions of $\gamma$ I've seen before, though seeing $\ln(n)$ and $1/n$ do remind me of the traditional way the constant is defined. I'm not sure how to get $b$ to drop out either. Why does $\zeta(s)$ appear to have this symmetry around $\gamma$? Thank you
EDIT: If your looking at this thread in the future someone gave me a solution here