$\lim_{\sigma \to \infty} \arg(\zeta(\sigma +iT)) = ?$ 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
$$
\lim_{\sigma \to \infty} \arg(\zeta(\sigma +iT)) = 0.
$$
 A: 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 $0$.
