I've been working on some real analysis homework and I have been tasked with finding an example where a sequence $(a_n)$ is unbounded, another sequence $(b_n)$ converges to zero, and $(a_nb_n)$ is not convergent. I settled on $a_n=n^2$ and $b_n=\tfrac{1}{n}$, where $n \in \mathbb{N}$. This made me think though… as even though $n^2$ diverges to infinity as $n$ increases, it has a lower bound… so is it still unbounded? I know that by definition, a bounded function is any function that is bounded both above and below, so does this mean that any function that does not fit this definition is necessarily unbounded?
Can an unbounded sequence have a lower bound and no upper bound (and vice versa)
real-analysissequences-and-series
Best Answer
If you want a sequence $a_n$ that doesn't have a lower bound, take $a_n = (-1)^n.n^2$ and $b_n = (-1)^n/n$. This should do the trick.