# alternative definitions for limit of a sequence

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.

• The constant $k$ is superfluous and $\le$ vs. $<$ doesn't make a difference. – Yves Daoust Aug 3 '19 at 15:07
• Yes but why? This means that there exists a "tacit property" that is still valid in the alternative versions. Which one? – zeroKnowl Aug 3 '19 at 15:11
• $\forall \epsilon$ makes $\le$ vs. $<$ irrelevant. – Yves Daoust Aug 3 '19 at 16:00

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 $$|a_n - l | for all $$n\geq N$$ and thus (1) is satisfied.

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