Let ${a_n}$, ${q_n}$ be sequences and ${a_n} \rightarrow a> 0$, ${q_n} \rightarrow +\infty$ proof ${a_n}\cdot {q_n}\rightarrow + \infty$

Question

I'm trying to use contradiction and the definitions of convergence and divergence, but I'm not sure of been solving anything

Answer 1

Let's first review the various definitions of limits we're going to need. We say $q_n\to\infty$ if and only if for all real numbers $K$, there exists some natural number $M$ such that $q_n>K$ whenever $n>M$. Furthermore, we say that $a_n\to a$ if and only if for all real $\varepsilon>0$, there exists some natural number $M$ such that $|a_n-a|<\varepsilon$ whenever $n>M$.

Now, we wish to show that $a_n q_m \to \infty$, so looking back at the above definition, we must be able to produce a natural number $M$ for every real number $K$ such that $a_nq_n >K$ whenever $n>M$. The key idea here is that we know that $q_n$ will eventually get arbitrarily large, and $a_n$ won't get too small, so we should be able to make their product arbitrarily large.

To make this precise, note that their must exist some number $M_1$ such that $q_n>\frac{2K}{a}$ whenever $n>M_1$ (you'll see why I've made this choice soon). Furthermore, there must exist some $M_2$ such that $a_n > a-a/2 = a/2>0$ whenever $n>M_2$ (I've just set $\varepsilon = a/2$ in the definition of the limit and rearranged a bit). Now, we want to only work in the setting when both of these inequalities are true, so set $M_3 = \mathrm{max}(M_1,M_2)$.

Now, if $n>M_3$, we have $a_nq_n > \frac{a}{2}\frac{2K}{a} = K$, thus $a_nq_n > K$, as desired.