Given $a_n \to \infty$ and $a_nb_n \to 2$, then $b_n \to 0$ I have bumped into some questions I wasn't too sure about, would appreciate your help.
it goes like this: given $a_n \to \infty$ and $a_nb_n \to 2$, then $b_n \to 0$
So I thought about a satisfying example $b_n = \frac{2}{n}$ $a_n \to n$, and I couldn't find a contradicting example, so I tried to proof it by formal way, but without any success.
this is what I tried:
let's assume $b_n \not\to 0$, thus either $b_n$ tends to some specific number K or it tends to $\pm\infty$
1.if it tends to some specific number K then $\lim_{n\to \infty} a_nb_n \to a_nK \to \pm\infty$ - Contradiction


*if $b_n$ tends to $+\infty$ then by "airithmetic of limits" the multiplication of series will also tend to $+\infty$ also Contradiction
3.however, in the case of $b_n \to -\infty$ I got stuck.
 A: You do not need a proof by contradiction, you can prove it directly.
The limit of $a_nb_n$ is 2, so from a certain index onwards it is between 1 and 3. Without loss of generality we can assume that it is always between 1 and 3 (otherwise cut off equal-length pieces from the front of all three sequences, which does not alter their convergence properties). Similarly assume that all the $a_n$ are positive.
Now $b_n=\frac{a_nb_n}{a_n},$ so $|b_n|\leq\frac3{a_n}$ which must converge to 0 because the numerator is constant and the denominator goes to infinity.
A: $b_n=a_nb_n a_n^{-1}\rightarrow 2\times 0=0$  
A: You are on the right track, but don't forget about the case where $b_n$ does not converge to number (think $(-1)^n$). What do you know when you assume $b_n\not\to0$? For every $\epsilon$ there are infinitely many terms of $b_n$ that are not within $\epsilon$ of zero. But since $a_n\to\infty$, from some point on all terms are at least $\frac{3}{\epsilon}$, so there are infinitely many terms of $a_nb_n$ that are at least $3$. Contradicting that $a_nb_n\to2$.
