The Riemann zeta function is described by the series $\zeta(s)=\sum 1/n^s$ for $\sigma>1$ where $s=\sigma+it$.
To show that the $\zeta(s)\rightarrow1$ as $\sigma\rightarrow\infty$ we might be tempted to use
$$\zeta(s)=\frac{1}{1^{s}}+\frac{1}{2^{s}}+\frac{1}{3^{s}}+\frac{1}{4^{s}}+\ldots$$
and say that as $\sigma\rightarrow\infty$, the magnitude of each term $|n^{-s}|=n^{-\sigma}$ tends to zero as $\sigma\rightarrow\infty$ for all n except $n=1$.
However - this assumes $\lim_{\sigma\rightarrow\infty}\sum 1/n^s = \sum \lim_{\sigma\rightarrow\infty} 1n^s$ which we haven't shown is true.
To do this we can use Tannery's Theorem (wiki, other).
Question: This question is about checking I have applied Tannery's theorem correctly.
Step 1
We start by stating Tannery's theorem to make sure we understand it. We start by defining:
$$S_j = \sum_{k}^{\infty} f_k(j) \; \text{ converges for all } j$$ $$ \lim_{j \rightarrow \infty} f_k(j) = f_k \; \text{ exists} $$
If the following bounds apply,
$$| f_k(j) | \leq M_k$$ $$ \sum_{k=0}^{\infty} M_k < \infty $$
Then we can say,
$$ \lim_{j \rightarrow \infty} S_j= \sum_{k=0}^{\infty} f_k $$
Step 2
We cast our challenge, and take care with the changed variable names.
We are interested in
$$ \lim_{\sigma\rightarrow\infty} \sum_n 1/n^s = \lim_{\sigma\rightarrow\infty} \sum_n f_n(\sigma) $$
where $f_n(\sigma) = 1/n^s = 1/n^{\sigma+it}$.
We then establish the preliminary conditions:
$$\sum_{n}^{\infty} f_n(\sigma) \; \text{ converges for all } \sigma>1$$
We are only interested in $\sigma>1$.
$$ \lim_{\sigma \rightarrow \infty} f_n(\sigma) = f_n \; \text{ exists} $$
Here:
$$ f_n = \begin{cases} 0 \text{ for } n>1\\ \\1 \text{ for } n=0 \end{cases} $$
Step 3
We then check the bounds:
$$| f_n(\sigma) | \leq M_n = 1/n^\sigma \text {converges for } \sigma > 1$$
$$ \sum M_n < \infty \text { for } \sigma >1$$
Step 4
Which finally allows us to say
$$ \lim_{\sigma \rightarrow \infty} \sum_n 1/n^s = \sum_{n} f_n = \sum 1+0+0+0 \ldots = 1 $$
This is close. There's really just one issue you need to fix: $M_n$ cannot depend on $\sigma$.
A simple way to fix this is to assume $\sigma \geq 2$ and then define $M_n = 1/n^2$.