How can I derive this expression related to the triangle inequality? In my real analysis text book, there is an expression like this for proof of some limit of a sequence:
$$n > N \implies \ |s_n - s | < 1$$
By using the triangular inequality, $n>N \implies \ |s_n| < |s| + 1$.
I don't understand how this can be.
 A: Let $N$ be a normed vector space over the field $\Bbb F$, where $\Bbb F = \Bbb R$ or $\Bbb F = \Bbb C$.  We denote the norm of $n$ by $\Vert n \Vert$ for $n \in N$.  Then for $m, n \in N$ we certainly have
$(n - m) + m = n, \tag{1}$
so by the triangle inequality
$\Vert n \Vert \le \Vert n - m \Vert + \Vert m \Vert, \tag{2}$
and reversing the roles of $m$ and $n$ we find that
$\Vert m \Vert \le \Vert n - m \Vert + \Vert n \Vert. \tag{3}$
(2) yields
$\Vert n \Vert  - \Vert m \Vert  \le \Vert n - m\Vert, \tag{4}$
and (3) gives
$\Vert m \Vert - \Vert n \Vert \le \Vert n - m \Vert. \tag{5}$
Taken together, (4) and (5) imply
$\vert \Vert n \Vert - \Vert m \Vert \vert \le \Vert n - m \Vert, \tag{6}$,
where $\vert c \vert$ is the ordinary absolute value for $c \in \Bbb F$.  Applying (6) to the inequality
$\vert s_n - s \vert < 1 \tag{7}$
we see that
$\vert \vert s_n \vert - \vert s \vert \vert < 1 \tag{8}$
as well, whence
$-1 < \vert s_n \vert - \vert s \vert < 1, \tag{9}$
from which 
$\vert s_n \vert < \vert s \vert + 1 \tag{10}$
immediately follows.  
I should in all honesty say I didn't invent (6); it is found in any of about a gajazillion analysis texts, under discussions about properties of norms.
Hope this helps.  Cheers, and as always,
Fiat Lux!
