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Assume $\lim(a_nb_n) =\infty$.

Approve or disapprove:

if $ 0 < b_n < a_n$ for almost every $n$, then $\lim a_n=\infty$

This is true, but I think I can disapprove it with this $a_n,b_n$: \begin{cases} a_n = 2n \:\text{ if }n > 5, & \text{otherwise: } 1/n \\ b_n = n\:\text{ if }n > 5, & \text{otherwise: } n ^ 2 \end{cases} Where am I wrong?

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Since $0 < b_n < a_n $ for almost all $n$, i.e. for all $n > m$ for some fixed $m$, you know that $a_n > \sqrt{a_n b_n}$ for $n > m$, and since $\lim{a_n b_n} = \infty$, so is its squareroot and hence so is $a_n$

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I'm assuming that by "for almost every $n$" you mean the following more rigorous property, and that $(a_n)_{n\in\mathbb{N}}, (b_n)_{n\in\mathbb{N}}$ are sequences of real numbers:

If there exists an $m\in\mathbb{N}$ such that for every $n>m$ it follows that $0<b_n<a_n$, and $\lim_{n\to\infty} a_nb_n = \infty$, then $\lim_{n\to\infty} a_n = \infty$.

As a comment has pointed out, your choice of $a_n$ and $b_n$ doesn't disprove the theorem, as $\lim_{n\to\infty} 2n = \infty$.

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  • $\begingroup$ Your counterexample doesn't obey $\lim{a_n b_n} = \infty$. $\endgroup$ – Sebastian Schulz Feb 8 at 14:06
  • $\begingroup$ My bad. I replied when the OP was editing his question to MathJax formatting and missed that part of the theorem. Thanks for the correction, I will edit my answer now. $\endgroup$ – zhantyzgz Feb 8 at 14:47

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