Lets $a_n$ and $b_n$ be two sequence that $0\leqslant a_n \leqslant 1$ and $0\leqslant b_n\leqslant 1$ for each $n$.
Prove or give a counterexample that:
a) if $\displaystyle\liminf_{n\to\infty} a_nb_n = 1$ then $\displaystyle\lim_{n\to\infty} a_n = 1$
b) if $\displaystyle\limsup_{n\to\infty} a_nb_n = 0$ then $\displaystyle\lim_{n\to\infty} a_n = 0$
I was trying to find a counterexample for hours and it's not working, I'd like to get a tip or anything...
(a) Clearly, we have : $0 \leqslant a_nb_n \leqslant a_n \leqslant 1$. Therefore, $$1 = \displaystyle\liminf_{n\to\infty} a_nb_n \leqslant \displaystyle\liminf_{n\to\infty} a_n \leqslant 1$$ implying : $\displaystyle\liminf_{n\to\infty} a_n= 1$
Again, $$1 \geqslant \displaystyle\limsup_{n\to\infty} a_n \geqslant \liminf_{n\to\infty} a_n = 1$$ implying : $\displaystyle\limsup_{n\to\infty} a_n = \liminf_{n\to\infty} a_n = 1$
Hence, we must have $\displaystyle \lim_{n\to\infty} a_n = 1$ .
(b) As per J. W. Tanner's comment above, take $\{a_n\} = (1 , 0 , 1 , 0 , \cdots)$ and take $\{b_n\} = (0 , 1 , 0 , 1 , \cdots)$ .
Then, $\displaystyle\limsup_{n\to\infty} a_nb_n = 0$ , but $\displaystyle\lim_{n\to\infty} a_n$ does NOT exist.