$\lim_{\sigma \to \infty} \arg(\zeta(\sigma +iT)) = ?$

Question

We know that for a real number $s$, $$ \lim_{s \to \infty} \zeta(s) = 1. $$

Now, I want to show that for a given fixed positive $T$, we have that $$ \lim_{\sigma \to \infty} \arg(\zeta(\sigma +iT)) = 0. $$

Answer 1

Suppose that $\sigma>1$. By the Euler product representation \begin{align*} &\left| {\arg \zeta (\sigma + iT)} \right| = \left| {\sum\limits_p {\arg \left( {1 - \frac{1}{{p^{\sigma + iT} }}} \right)} } \right| \\ &= \left| {\sum\limits_p {\arg \left( {1 - \frac{{\cos (T\log p)}}{{p^\sigma }} + i\frac{{\sin (T\log p)}}{{p^\sigma }}} \right)} } \right| \\ & = \left| {\sum\limits_p {\arctan \left( {\frac{{\sin (T\log p)}}{{p^\sigma - \cos (T\log p)}}} \right)} } \right| \\ &\le \sum\limits_p {\arctan \left( {\frac{1}{{p^\sigma - 1}}} \right)} \le \sum\limits_p {\frac{1}{{p^\sigma - 1}}} , \end{align*} where the sum runs over the primes. Letting $\sigma \to +\infty$ yields the limit of $0$.

Alternatively, $$ \left| {\zeta (\sigma + iT) - 1} \right| = \left| {\sum\limits_{n = 2}^\infty {\frac{1}{{n^{\sigma + iT} }}} } \right| \le \sum\limits_{n = 2}^\infty {\frac{1}{{n^\sigma }}} = \zeta (\sigma ) - 1, $$ whence $\zeta (\sigma + iT) \to 1$ as $\sigma \to +\infty$. Consequently, the phase of $\zeta (\sigma + iT)$ must tend to $0$.