# if $\lim (a_nb_n) = \infty$ and $0<b_n<a_n$ for almost every $n$, then $\lim a_n =\infty$

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?

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 $$a_n$$
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 $$0, and $$\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$$.
• Your counterexample doesn't obey $\lim{a_n b_n} = \infty$. – Sebastian Schulz Feb 8 at 14:06