What is the geometric interpretation of this limit :$\lim_{n\to {+\infty}}\zeta(n)+\zeta(\frac1n)=\frac12$?

Question

it is easy to show that $\lim_{n\to {+\infty}}\zeta(n)+\zeta(\frac1n)=\frac12$ , The latter expressed the relationship between $\zeta(n) $ and $\zeta(\frac1n)$ with $n$ is a positive integer , Then i have two questions here : The first I want to know here how i can compute $\zeta(n) $ using $\zeta(\frac1n)$ ?and the second What is the geometric interpretation of this limit :$\lim_{n\to {+\infty}}\zeta(n)+\zeta(\frac1n)=\frac12$ ?

Answer 1

There is no deep underlying geometric picture. The fact that $\zeta(n) \to 1$ as $n \to \infty$ is nothing more than the fact that the first term is $1$, and every other term decays. Similarly, $\zeta(x)$ is continuous at $x = 0$, and $\zeta(0) = -1/2$. For two quite simple reasons, we see that $\zeta(n) + \zeta(1/n) \to 1 - \frac{1}{2} = \frac{1}{2}$.

It may be beneficial to ruin some of the artificial symmetry. It is also true that $$ \lim_{n \to \infty} \zeta(3n) + \zeta(\tfrac{1}{7n}) = \frac{1}{2}.$$ Or instead of $3$ and $7$, you can use any positive real numbers you want. I think the complete freedom in this choice shows how unrelated $\zeta(n)$ and $\zeta(1/m)$ really are.

Answer 2

I. The Graph of $\zeta(s)$ for only real numbers (source)

We see :

• $2$ asymptotes to the $x=1$: $y_1=0$ and $y_2=1$
• for $x\in[0;1)$ : $\zeta(x)\leq-\frac12$

II. (Intuitive) Let's define the function $g(x)$: \begin{cases} \frac12 & \quad \text{ $x=0$}\\ \ \zeta(x)+\zeta(\frac1x) & \quad \ \text{ $x>0 \land x\neq1$}\\ 0 & \quad \text{ $x=1$} \end{cases}

For $x\rightarrow \infty:g(x)$ going closer to the asyptote $y=\frac12$ ( which is the center axis between $y_1$ & $y_2$)

Here some points I 'calcule' ( w/ WolframAlpha plus Wikipedia help)