Undergrad How Is Divergence to Infinity Defined in Contrast to General Divergence?

[Mr Davis 97]

The general definition for a sequence to diverge is the negation of what it means for a sequence to converge: $\forall L\in\mathbb{R}~\exists\epsilon>0~\forall N\in\mathbb{N}~\exists n\ge N$, $|a_n - L| \ge \epsilon$. How does this general definition of divergence relate to the definition of a sequence diverging specifically to infinity, which is $\forall M \in \mathbb{R} ~ \exists N \in \mathbb{N} \forall n \ge N$, $a_n > M$?

 

[andrewkirk]

The second is a sub-case of the first, so if the second criteria is met, so will the first be. That needs to be proved. It is not difficult. The reverse implication does not hold. Oscillating sequences are a counterexample.

 

[Mr Davis 97]

The second is a sub-case of the first, so if the second criteria is met, so will the first be. That needs to be proved. It is not difficult. The reverse implication does not hold. Oscillating sequences are a counterexample.

How would I go about showing that the second criteria implies the first, out of curiosity? I tried to take a stab at it, but I got buried by the quantifiers...

EDIT: Actually, maybe I see how to do. If we argue by contradiction we see that $a_n$ must converge, and so is bounded. But that contradicts the fact that it is unbounded...

 

[RPinPA]

Not quite formally: Let L be any real number and M any number M > L. Define $M - L = \epsilon$. Then eventually all $a_n > M$ which certainly meets the condition $|a_n - L| \geq \epsilon$