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Let $(a_n),(b_n)$ sequences such that $\lim_{n\rightarrow \infty} a_n b_n=1$

Prove that if $\forall n\in \mathbb{N}, 0\leq a_n,b_n\leq 1$ then $\lim_{n\rightarrow \infty} a_n=\lim_{n\rightarrow \infty} b_n=1$

My attempt (which may suck as the semester didn't start yet so I'm studying on my own):

The problem here is we don't know if the sequences converge, I'm pretty sure I need to resort to sub-sequences.

By the theorem every sequence has a converging subsequence, there are $(\alpha_k),(\beta_k)$ such that $(a_{\alpha_k}),(b_{\beta_k})$ converge, denote the limits $A,B$ respectively. Now we get $\lim_{k\rightarrow \infty} a_{\alpha_k}b_{\beta_k}=AB$ which is where I'm stuck, as I want to show this is the limit of $a_nb_n$. Even if I did show this, and after hard work (which I'm still not sure I can do on my own) I show $A,B=1$, I still don't know that the original sequences converge!

I feel like I took the completly wrong direction here. I really need some assistance, so any help is appreciated in advance.

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HINT

Use the definition and the fact that: $|a_n - 1| \le |a_nb_n -1| \lt \epsilon$ and $|b_n - 1| \le |a_nb_n -1| \lt \epsilon$ for $n \ge N$

$|a_n - 1| \le |a_nb_n -1|$ it's easy to prove because it is equivalent to $1-a_n \le 1-a_nb_n$

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Let $i=\liminf_{n\to\infty} a_n$, $s=\limsup_{n\to\infty} a_n$, $I=\liminf_{n\to\infty} b_n$, $S=\limsup_{n\to\infty} b_n$. Now if $i<1$ then $S>1$, a contradiction. If $I<1$ then $s>1$, a contradiction. So $i=I=s=S=1$ and the claim follows. The details are left for you to fill in.

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  • $\begingroup$ The other answer helped me already but I insist on understanding your answer aswell. I understand intuitively the things you claimed but how do you formally prove them? The rest I understand since if these limits are equal the original sequence converges to the same number. $\endgroup$ – Theorem Oct 17 '16 at 18:03

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