Definition of convergence to infinity Does the following definition hold true for every series that converges to infinity?
Definition- A sequence $x_n$ is said to converge to $\infty$ if, for every $a>0$ and $\epsilon>0$, there exists $N\in \mathbb{N}$, such that for $n>N$ it is true that $|x_n-a|>\epsilon$
Clarification: For example we can use this definition to determine that $x_n=\sqrt{n}$ converges to infinity. But if our sequence is $1,0,2,0,3,0,4,0,5...$ then the above definition can't be used.
Can this definition be modified so that oscillating sequences can also be tested for convergence to infinity?
 A: Given a sequence $(x_n)_n$, if for every $a>0$ and for every $\epsilon>0$, there exists $N\in \mathbb{N}$, such that, for $n>N$ it is true that $|x_n-a|>\epsilon$ then 
$$|x_n|+|a|\geq |x_n-a|>\epsilon\implies |x_n|>\epsilon-|a|$$
and therefore, by the arbitrarity of $a$ and $\epsilon$
$$\lim_{n\to +\infty}|x_n|=+\infty.$$
Also the other implication holds. More simply, a definition of $\lim_{n\to +\infty}|x_n|=+\infty$ should be: for every $a>0$ there exists $N\in \mathbb{N}$, such that, for $n>N$ it is true that $|x_n|>a$ (no need of the $\epsilon$ part).
The sequence $1,0,2,0,3,0,4,0,5, \dots$ has no limit and it does not converge to $+\infty$ (although it is unbounded). On the other hand, the oscillating sequence $1,-1,2,-2,3,-3,4,-4, \dots$ has no limit, it does not converge to $+\infty$ or $-\infty$ but it does satisfy your definition.
A: 
But if our sequence is $1,0,2,0,3,0,4,0,5...$ then the above definition can't be used.

Yes, it can. The definition can be used to come to a conclusion that the above sequence does not converge to $\infty$.


Can this definition be modified so that oscillating sequences can also be tested for convergence to infinity.

The definition already can be used to test for convergence of oscillating sequences to infinity. Using the definition, it can be proven that such sequences do not converge to $\infty$.
A: The definition is: a sequence $x_n$ is said to converging to $+\infty$, if 
$$\forall a\in \mathbb{R} \quad \exists N\in \mathbb{N}\quad n>N \implies x_n>a$$
Note that we don’t need to set $a>0$ and we don’t needi $\epsilon$ for the definition.
