# $\lim_{t \to \infty}\zeta(\frac{1}{2} + it)$

I am learning about to zeta-function, I am a beginner. I am trying to find:

$$\lim_{t \to \infty}\zeta(\frac{1}{2} + it)$$

From the book Theory of the Riemann zeta-function-clarendon by Titchmarsh in the theorem 8.12 I know:
If $$\frac{1}{2} \leq \sigma < 1$$ the $$|\zeta(\sigma + it)|> e^{\log^{\alpha}y}$$ with $$\alpha < 1- \sigma$$ and for some indefinitely large values of $$t$$

For my case if $$0<\alpha<\frac{1}{2}$$ then since $$|\zeta(1/2 + it)|> e^{\log^{\alpha}t}$$ for some indefinitely large values of $$t$$ since $$t \to \infty$$ then I don't know if that's enough to conclude that $$\lim_{t \to \infty}\zeta(\frac{1}{2} + it)$$ does not converge.

• What do you mean by $|\zeta(1/2 + it)|> \infty$? How can something be larger than infinity? For which values of $t$? We know that $|\zeta(1/2 + it)|$ has infinitely many zeros, so Titchmarsh's inequality cannot hold for all $t$. What is $t$ in Titchmarsh's inequality? You ahve $x$ and $y$ on the left-hand side. Please revise your question carefuly and address these issues.
– Gary
Nov 7, 2021 at 22:55
• Thanks you Mr Gary, I tryed to write better my doubt. Nov 7, 2021 at 23:39
• The theorem still makes no sense. What is the relation of $x$ and $y$ to $t$? As I said $\zeta(1/2+i t)$ is $0$ infinitely often as $t \to +\infty$, so you cannot bound it from below by a positive expression of $t$.
– Gary
Nov 7, 2021 at 23:39
• It's on page 204 of the second edition, for the record. Nov 7, 2021 at 23:46
• I corrected the statement. You replaces $\sigma$ and $t$ from the book by $x$ and $y$ for unknown reasons in certain places. Titchmarsh talks about $\limsup_{t\to +\infty}$. The limit does not exist, see the answer below.
– Gary
Nov 7, 2021 at 23:46

One very clear point is that it is known by now that there are infinitely-many zeros on the critical line (Levinson... Conrey... showed that at least 2/5 (?) or so are on-the line...), but/and away from zeros zeta grows. So there's no actual limit of $$\zeta({1\over 2}+it)$$ as $$t\to \infty$$.
Naturally, with or without RH, things are even more chaotic to the right of $$\Re(s)={1\over 2}$$. For example, Voronin's universality theorem overwhelmingly indicates that there are no elementary asymptotics on any vertical line to the right of $$\Re(s)={1\over 2}$$. Overkill, yes, but really decisive.
By Titchmarsh's theorem, $$\mathop {\lim \sup }\limits_{t \to + \infty } \,\left| {\zeta\! \left( {\tfrac{1}{2} + it} \right)} \right| = + \infty .$$ We also know that there is an infinite number of zeros along the critical line (with an accumulation point at infinity), so $$\mathop {\lim \inf }\limits_{t \to + \infty } \,\left| {\zeta\! \left( {\tfrac{1}{2} + it} \right)} \right| = 0.$$ Since the two are not equal, the limit $$\lim_{t \to + \infty } \left| {\zeta\! \left( {\tfrac{1}{2} + it} \right)} \right|$$ does not exist, and hence the limit $$\lim_{t \to + \infty } \zeta\! \left( {\tfrac{1}{2} + it} \right)$$ does not exist.