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

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 :$\lim_{n\to {+\infty}}\zeta(n)+\zeta(\frac1n)=\frac12$ ?

• – user547564 Jul 28 '18 at 21:35

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.

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)