2
$\begingroup$

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 $\lim_{n\to\infty}a_n b_n=\infty$. $\blacksquare$

$\endgroup$
1
  • 4
    $\begingroup$ Looks good. Might as well take $\alpha=1$ to simplify the algebra a tiny bit $\endgroup$
    – QC_QAOA
    Commented Aug 15, 2022 at 19:15

1 Answer 1

1
$\begingroup$

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).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .