# Some questions about proof for limit of sequence $a_n := 1 + \frac{1}{n}$ is 1

I'm trying to understand a proof for the limit of the sequence $$a_n := 1 + \frac{1}{n}$$ and $$n \in \mathbb{N}$$ which should be 1.

So, the proof which I have here starts with:

$$\left( \lim_{n \to \infty} a_n = 1 \right) :\Leftrightarrow \left( \forall \varepsilon \in \mathbb{N} \right) \left(\exists N \in \mathbb{N} \right)\left(\forall n \in \mathbb{N} \right) \left[ n > N \Rightarrow \left| a_n - a \right| < \varepsilon \right]$$

which, when including $$a_n := 1 + \frac{1}{n}$$ and $$a := 1$$ should results in :

$$\left| a_n - a \right| < \varepsilon \Leftrightarrow \left| \left( 1 + \frac{1}{n} \right) -1 \right| < \varepsilon \Leftrightarrow \frac{1}{n} < \varepsilon \Leftrightarrow n > \frac{1}{n}$$

Here I'm having a problem: I know that

$$\left| \left( 1 + \frac{1}{n} \right) -1 \right| < \varepsilon \Leftrightarrow -\varepsilon < \left( 1 + \frac{1}{n} \right) -1 < \varepsilon \Leftrightarrow -\varepsilon < \frac{1}{n} < \varepsilon$$ as $$\varepsilon$$ is per definition larger than 0, but why is $$-\varepsilon <$$ left out in the term above?

Thanks!

It is left out because there isn't an essence in checking $$-\varepsilon$$ as a lower bound. Note that you arrive at the expression :
$$\left| \frac{1}{n} \right| < \varepsilon \implies \frac{1}{n} < \varepsilon$$
That's because $$1/n >0 \; \forall n \in \mathbb N$$ and thus it is trivially larger than any negative number.
Because we already know that $$0<\frac1n$$, we get $$-\varepsilon<\frac1n$$ for free (and almost trivially), no matter what $$\varepsilon$$ and $$n$$ are. Equivalently, we know that $$\frac1n$$ is positive, so the absolute value signs may be dropped without any consequence.
Every negative number is less than every positive number so no purpose is served in writing $$-\epsilon <\frac 1 n$$.