The question of proving $\zeta(s)\rightarrow1$ as $\sigma\rightarrow +\infty$ has been posed several times.
The answers seem to be overly complex - eg (link).
My previous question (link) was not answered, but was challenged regarding the interchange of limit operators.
Here I attempt to use Tannery's Theorem (link) to justify the interchange of limit operators.
Here $\zeta(s)=\sum 1/n^s$ for complex $s=\sigma+it$, and $\sigma>1$.
Question: Is the following simple argument correct?
Tannery's Theorem says that
$$ \lim_{\sigma \rightarrow +\infty} \sum 1/n^s = \sum \lim_{\sigma \rightarrow +\infty} 1/n^s$$
if $|1/n^s| \leq M_n$ and $\sum M_n < \infty$.
In our case $|1/n^s| \leq 1/n^{\sigma}$ and $\sum 1/n^{\sigma} < \infty$ for $\sigma>1$. Thus, the conditions for Tannery's theorem are met.
Therefore $\zeta(s)\rightarrow1$ as $\sigma\rightarrow +\infty$.