Traditionally the limit of a sequence is defined as $\lim_{n\to\infty} x_n = L \Leftrightarrow$
$$(\forall \varepsilon > 0)(\exists N > 0)(\forall n \in \mathbf{N})(n \geq N \Rightarrow |x_n-L|<\varepsilon). \tag{1} $$
However, sometimes it seems more helpful to use the following formulation.
$$(\forall \varepsilon > 0)(\exists N > 0)(\forall n \in \mathbf{N}_{\geq N})(|x_n-L|<\varepsilon) \tag{2} $$
I’m not entirely sure that both statements are exactly equivalent, but I do have some reasons to believe so. Considering $(1)$ is immediately true $\forall n \in \mathbf{N}_{<N}$ because of the implication with a false antecedent and that for bigger $n$ the statement is true if $|x_n-a|<\varepsilon$, it seems like they are exactly equivalent.
This formulation would allow for a slightly nicer negation, but as I said, I’m not convinced enough to use $(2)$ on an exam. Could someone explain to me why there is indeed a devil in the details or why they are exactly the same?
Thanks in advance.
Let $P(n)$ abbreviate $|x_n-L|<\varepsilon$ . We have to show that the statements
$$\forall n \in \mathbf{N} : n \geq N \Rightarrow P(n). \tag{1} $$ and $$\forall n \in \mathbf{N}_{\geq N} : P(n) \tag{2} $$ are equivalent.
$(1) \implies (2)$:
Let $n \in \mathbf{N}_{\geq N}$. We have to show that $$P(n)$$ is true.
Since $n \in \mathbf{N}$, we know that the implication $$n \geq N \Rightarrow P(n)$$ is true. Since the implication's premise $n \geq N$ is true for $n \in \mathbf{N}_{\geq N}$, we see that the implication's conclusion $$P(n)$$ must be true.
$(2) \implies (1)$:
Let $n \in \mathbf{N}$. We have to show that the implication $$n \geq N \Rightarrow P(n)$$ is true.
Case 1. $n \ge N$. Then by $(2)$ the implication's conclusion $$P(n)$$ is true. Thus the implication is true for this $n$.
Case 2. $n < N$. Then the implication's premise $$n \ge N$$ is false. Thus the implication is true for this $n$.