Prove using the Sandwich Theorem for sequences

Question

CLAIM: Prove with using the Sandwich Theorem that if $a_n$ convergent and $b_n→∞$ then $\frac{a_n}{b_n}→0$

DO NOT USE: (Bounded $X$ $d_n→0$) $→0$

Answer 1

Since ${a_n}$, there is a real constant number $A>0$ such that $$ |a_n| \le A \quad \forall n. $$ Also, because $\lim b_n=\infty$, there is an integer $N$ such that $$ b_n>0 \quad \forall n\ge N. $$ Hence, for all $n\ge N$ we have $$ -\frac{A}{b_n} \le \frac{a_n}{b_n} \le \frac{A}{b_n}, $$ and therefore by the Sandwich Theorem we have $$ \lim_{n\to \infty}\frac{a_n}{b_n}=0. $$

Answer 2

Hint:

A convergent sequence is bounded, and $$\frac{a_n}{b_n}\to 0\iff \biggl\lvert\frac{a_n}{b_n}\biggr\rvert\to 0.$$

Answer 3

If $(a_n)$ converges to $a$, then eventually we get the inequality: $$\frac{a-1}{b_n}\le\frac{a_n}{b_n}\le\frac{a+1}{b_n}.$$ Both "sandwiching" terms are of the form $(c/b_n)$ for $c$ constant and $(b_n)$ going to infinity, so they they converge to 0.