In the definition of sequences diverging to infinity, why must the constants be positive?
We are given the following definitions
A sequence $(a_n)$ diverges to $\infty$ if for each $ M \in \mathbb{R}^+ \exists N_M \in \mathbb{N} \ \text{such that } \\ a_n > M \ \forall n \geq N_M$
A sequence $(a_n)$ diverges to $- \infty$ if for each $m \in \mathbb{R}^+ \exists N_m \in \mathbb{N} \ \text{such that} \\ a_n < m \ \forall n\geq N_m$
I understand what the definition implies, the only thing that I do not feel certain about, is why must M and m be greater than zero?
EDIT: Here is the exact extract from my notes
$\endgroup$ 43.3.8 $\quad$ Definition
(a) $\quad$ A sequence $(a_n)$ diverges to $\infty$ (Write: $\lim\limits_{n\to\infty}a_n=\infty$)
if for each real number $M\gt0$ there exists a $N_M\in\mathbb N$ such that $a_n\gt M$ for all $n\geqslant N_M$.(b) $\quad$ sequence $(a_n)$ diverges to $-\infty$ (Write: $\lim\limits_{n\to\infty}a_n=-\infty$)
if for each real number $m\gt0$ there exists a $N_m\in\mathbb N$ such that $a_n\lt m$ for all $n\geqslant N_m$.
1 Answer
$\begingroup$What this definition means is that a sequence $(a_n)_{n\in\mathbb{N}}$ diverges to $\infty$ if, and only if, for any "threshold" $M$ of your choosing, after a while* all the elements of the sequence are above the threshold**. That is, no matter how high a threshold you fix, the sequence will eventually reach it, and stay above it afterwards.
The choice $M\geq 0$ is completely arbitrary: the definition remains exactly the same if you allow $M\in\mathbb{R}$ (if you choose a threshold $M < 0$, then you can take $N_M=N_0$).
*a while: $\exists N_M$
** $\forall n\geq N_M$, $a_n \geq M$
$\endgroup$ 1More in general
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