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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!

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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.

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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.

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Every negative number is less than every positive number so no purpose is served in writing $-\epsilon <\frac 1 n$.

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