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?
Best Answer
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.