real analysis - Show that if $|s_{mn}-S|

This is the exercise 2.8.5 of the book Understanding analysis of Abbott. To put you in context I have that

1. The sequence of partial sums $(s_{mn})$ is absolutely convergent, and the definition is $s_{mn}=\sum_{i=1}^{m}\sum_{j=1}^{n}a_{ij}$. All the next statements are derived from here.




2. $(s_{nn})\to S$




3. $|s_{mn}-S|<\varepsilon,\forall n>m\ge N$




4. $\lim_{n\to\infty} s_{mn}=r_1+r_2+\cdots+r_m$

Now it is supposed that I can show that

$$|s_{mn}-S|<\varepsilon, \forall n>m\ge N\implies |(r_1+r_2+\cdots+r_m)-S|\le\varepsilon$$

Maybe this is a very dumb question but I cant see where the "$\le$" comes. So my thought is that the "$\le$" symbol must appear as a consequence of some order relation between limits but I cant see how, can you enlighten me please? Thank you in advance.

EDIT: to be specific, I cant see why the case is not just

$$|s_{mn}-S|<\varepsilon, \forall n>m\ge N\implies |(r_1+r_2+\cdots+r_m)-S|<\varepsilon$$

instead of the stated one. To me if $|s_{mn}-S|<\varepsilon$ holds for any $n>m\ge N$ I cant see why in the limit when $n\to\infty$ this maybe different.

Answer

If for some sequence $(x_n)$ we have $x_n < \epsilon$ for all $n$ and $x_n \to x$ then we can conclude $x \leqslant \epsilon$. This is easily proved by contradiction assuming $x > \epsilon$. Then we arrive at a contradiction where infinitely many $x_n$ are in a neighborhood $(x- \delta,x+\delta)$ with $x_n > x - \delta > \epsilon$.

The inequality need not be strict to allow for the possibility that $\epsilon$ is actually the limit. For example, $x_n = 1 - 1/n < 1$ for all $n$ but $\lim_n (1 - 1/n) = 1.$

In your case, the sequence in question is $x_n = |s_{mn}-S|$ and $\lim_n s_{mn} = |(r_1 + \ldots + r_m) - L|.$