# Suppose $\lim_{n\to\infty}a_n=\infty$ and $\lim_{n\to\infty}b_n=\infty$. Prove $\lim_{n\to\infty}a_nb_n=\infty$.

I'm trying to prove Theorem 2.3.6(b) from Introduction to Mathematical Analysis I (Lafferriere, Lafferriere, Nguyen) as practice. The theorem says:

Let $$\{a_n\}$$ and $$\{b_n\}$$ be sequences of real numbers. Suppose $$\lim_{n\to\infty}a_n=\infty$$ and $$\lim_{n\to\infty}b_n=\infty$$. Then $$\lim_{n\to\infty}a_nb_n=\infty$$.

Their definition of sequences that diverge to $$\infty$$ is:

A sequence $$\{a_n\}$$ is said to diverge to $$\infty$$ if for every $$M\in\mathbb{R},$$ there exists $$N\in\mathbb{N}$$ such that $$a_n>M$$ for all $$n\geq N$$. In this case, we write $$\lim_{n\to\infty}a_n=\infty$$.

Could anyone verify whether my proof is correct?

Proof: Fix $$\alpha>0.$$ Since $$\lim_{n\to\infty}a_n=\infty$$, there exists some $$N_1\in\mathbb{N}$$ such that $$a_n>\alpha$$ for all $$n\geq N_1.$$ Now suppose $$M\in\mathbb{R}.$$ Then since $$\lim_{n\to\infty}b_n=\infty,$$ there exists some $$N_2\in\mathbb{N}$$ such that $$b_n>M/\alpha$$ for all $$N\geq N_2.$$ Choose $$N=\max\{N_1,N_2\}$$. Then we now have that $$a_nb_n>\alpha(M/\alpha)=M$$ for all $$n\geq N$$. Since $$M\in\mathbb{R}$$ is arbitrary, we conclude $$\lim_{n\to\infty}a_n b_n=\infty$$. $$\blacksquare$$

• Looks good. Might as well take $\alpha=1$ to simplify the algebra a tiny bit Aug 15, 2022 at 19:15

The proof is correct. You can simply see that $$a_n$$ is bounded from below for large enough $$n$$ (take $$\alpha>0$$ to be the lower bound), so $$a_n b_n>\alpha b_n\to\infty$$ (as $$b_n$$ is positive for $$n$$ large enough).