$a_nb_n\rightarrow c$ and $b_n\rightarrow 0$ for $n\rightarrow\infty$ implies $(a_n)_{n\in\mathbb{N}}$ unbounded? Let $c\in\mathbb{R}\backslash\{0\}$, $(a_n)_{n\in\mathbb{N}}$ and $(b_n)_{n\in\mathbb{N}}$ sequences with values in $\mathbb{R}$, such that $a_nb_n\rightarrow c$ and $b_n\rightarrow 0$ for $n\rightarrow\infty$, where each $b_n>0$. In fact, $b_n$ can be written as $b_n=n^{-1} d$ for some constant $d>0$.
Does this imply, that $(a_n)_{n\in\mathbb{N}}$ is unbounded? I think this must be true, because suppose the sequence is bounded, let's say by a constant $C>0$, we find that $0\le |a_n||b_n|\le C|b_n|\rightarrow 0$ for $n\rightarrow\infty$, which would be a contradiction to $c\ne 0$.
Is this proof right?
 A: A handy result in sequences is "bounded times null is null." Formally:

If $x_n$ is bounded, and if $y_n \to 0$, then $x_n y_n \to 0.$

Now assume that $b_n$ is null. The result just given tells us that if $(a_n b_n \to 0)$ does not hold, then it must be that $a_n$ is unbounded.
Your proof is essentially a proof of this, with the slight difference that you have assumed that $a_n b_n$ converges. The assumption "$a_n b_n$ converges to a non-zero $c$" is a special case of "$(a_n b_n \to 0)$ does not hold." (The essence is that we can know $|b_n|<\epsilon/C$ for big indices.)
In your first paragraph, you consider some other possible additional assumptions. We have seen already that "$b_n$ strictly positive for all $n$" is not needed to conclude that $a_n$ is unbounded, whether of the form $d/n,\,d>0$ or not.
If you are willing to assume $a_n b_n \to c>0$ and also $b_n>0$ holds always, then $a_n$ is ultimately positive, and $1/a_n \to 0$.
But as Mark Viola has pointed out, if $c<0$ then $a_n \to +\infty$ may not hold. If you want a theorem that is close to your original title, you can prove:

$a_nb_n\rightarrow c\not =0$ and $0\not =b_n\rightarrow 0$ for $n\rightarrow\infty$ implies $|a_n|\rightarrow\infty$ for $n\rightarrow\infty$.

