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In several proofs I noticed that authors consider slightly different inequalities to prove that a sequence $(a_n)$ converges to a limit $l$, for example:

$$\forall \epsilon>0 \: \exists N \: \forall n \ge N \: |a_n - l | \le \epsilon$$ and $$\forall \epsilon>0 \: \exists N \: \forall n \ge N \: |a_n - l | < k\epsilon$$ where k is a constant.

The aforementioned versions are different from the following traditional definition: $$\forall \epsilon>0 \: \exists N \: \forall n \ge N \: |a_n - l | < \epsilon$$

Why can we consider them as equivalent?

Thanks a lot.

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  • $\begingroup$ The constant $k$ is superfluous and $\le$ vs. $<$ doesn't make a difference. $\endgroup$ – Yves Daoust Aug 3 '19 at 15:07
  • $\begingroup$ Yes but why? This means that there exists a "tacit property" that is still valid in the alternative versions. Which one? $\endgroup$ – zeroKnowl Aug 3 '19 at 15:11
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    $\begingroup$ $\forall \epsilon$ makes $\le$ vs. $<$ irrelevant. $\endgroup$ – Yves Daoust Aug 3 '19 at 16:00
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The point is that $\epsilon$ can be chosen to be arbitrarily small. Just using this fact, I will show that all three definitions are equivalent:

Let $(a_n)$ be a sequence and $\ell$ a number. Consider the following assertions:
(1) $\forall \epsilon>0 \: \exists N \: \forall n \ge N \: |a_n - l | \le \epsilon$
(2) $\forall \hat\epsilon>0 \: \exists \hat N \: \forall n \ge \hat N \: |a_n - l | < \hat \epsilon$
(3) $\forall \tilde\epsilon>0 \: \exists \tilde N \: \forall n \ge \tilde N \: |a_n - l | < k\tilde\epsilon$ where $k>0$ is a constant.

Claim: Assertions (1), (2), (3) are equivalent.

Proof:
(1) $\implies$ (2) Let $\hat \epsilon >0$ and set $\epsilon = \hat\epsilon/2$. Then $0<\epsilon<\hat \epsilon$ and, by (1), there exists $N$ such that $|a_n - l | \le \epsilon$ for all $n\geq N$. Set $\hat N = N$, then it holds $|a_n - l | \le \epsilon < \hat \epsilon$ for all $n\geq \hat N$ and thus (2) is satisfied.

(2) $\implies$ (3) Let $\tilde \epsilon >0$ and set $\hat \epsilon = k\tilde \epsilon$. Then $\hat \epsilon>0$ and, by (2), there exists $\hat N$ such that $|a_n - l | \le \epsilon$ for all $n\geq \hat N$. Set $\tilde N = \hat N$, then it holds $|a_n - l | <\hat\epsilon = k\tilde \epsilon$ for all $n\geq \tilde N$ and thus (3) is satisfied.

(3) $\implies$ (1) Let $\epsilon >0$ and set $\tilde \epsilon = \epsilon/k$. Then $0<\tilde \epsilon$ and $k\tilde\epsilon \leq \epsilon$. By (3), there exists $\tilde N$ such that $|a_n - l | < k\tilde\epsilon$ for all $n\geq \tilde N$. Set $N= \tilde N$, then it holds $|a_n - l | <k\hat\epsilon \leq \epsilon$ for all $n\geq N$ and thus (1) is satisfied.

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    $\begingroup$ Very clear, thanks a lot for your time! $\endgroup$ – zeroKnowl Aug 6 '19 at 17:46

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