Sequence converges means it is bounded above and below

Question

What happens in a situation where $\{a_n\}$ has an asymptote, will there ever be a case? Because if there is one then it's no longer bounded.

Answer 1

A sequence is function from $\mathbb N\to X$. The domain of the sequence (the set of indexes) is the natural numbers. It does not make sense to talk of a vertical asymptote if its domain is the natural numbers.

Intuitively to say $f$ has a vertical asymptote at $a$ means that at values "closer and closer" to $a$ that $f(x)$ gets larger and larger. But if our domain is the natural numbers or integers, taking values "closer and closer" to $a$ just doesn't make sense.

Formally....

A function $f:\mathbb R \to \mathbb R$ has a vertical asymptote if $\lim_{x\to a^-} f(x) = \infty$.

So if $f(n) =a_n$ so $f:\mathbb N \to \mathbb R$, what does it mean to say $\lim_{n\to a^-} f(x) = \infty$.

By definition it means for any $M$ there is an small number $\delta$ so that if $a-\delta < x < a$ then $f(x) > M$.

That can not be true. Let $a-1 \le n <a$. And let $M > f(n)$.

We can't have $\delta > a-n$ because then we'd have $a-\delta < n < a$ but $f(n) < M$.

And we can't have $\delta \le a-n$ because that would mean, if $a-\delta < x < a$ then $x\not \in \mathbb Z$. So $f(x)$ is not defined and it is not true that $f(x) > M$.

Answer 2

If the sequence is convergent i.e., $\lim s_n = s$ it has a horizontal asymptote $y=s$. Convergent sequence are bounded.

If the asymptote is not horizontal, the sequence is not convergent.