0
$\begingroup$

I got this question :

Consider $a_n,\:b_{n\:}$ sequences such that for every n , $0\le a_n\le \:b_{n\:}$.

Let $\lim _{n\to \infty }\left(\frac{b_n}{a_n}\right)\:=\:1$, and $a_n$ is a bounded sequence.

Prove that $\left(a_n-b_n\right)_{n=1}^{\infty \:}\:\rightarrow \:0$.

I tried little bit by myself to understand what is given:

By definition of limit, we see that : $\forall\epsilon, \exists n_{0} \forall n> n_{0}\Rightarrow |\frac{b_{n}}{a_{n}}- 1|< \varepsilon $, and also that exist some $M$ that for every $n$, $-M<a_n<M$.

Now i'm looking for that right?: i need to prove that $ \forall\epsilon, \exists n_{0} \forall n> n_{0}\Rightarrow |a_{n}-b_{n}- 0|< \varepsilon $

Here is where i struggle: i choose some $\epsilon$. What i need to find? an $N2$ that for every $n>N2, |(a_{n}-b_{n})- 0|< \varepsilon$ ?

How to start?

$\endgroup$
0
$\begingroup$

Hint: use may the fact that $|a_n-b_n|=|a_n||\frac{b_n}{a_n}-1|$ in your proof .

$\endgroup$
  • $\begingroup$ Yes, but what i actually need to write to do it formal? $\endgroup$ – user2637293 Sep 18 '14 at 1:09
  • $\begingroup$ I would recommend you to fill in the gap yourself. but if not interested you can look at the proof provided by @Ishfaaq down here. $\endgroup$ – BigM Sep 18 '14 at 1:12
0
$\begingroup$

Hints

First of all notice that $\left|{\dfrac{b_n}{a_n} - 1}\right| = \dfrac{|(b_n - a_n) - 0|}{|a_n|} $

Secondly note that $ |a_n| \le M \;\; \forall n \in \Bbb N$

Now let $\epsilon \gt 0$ be arbitrary and think what you can do with a quantity like $$ \dfrac{\epsilon}{M} \gt 0 $$

UPDATE: Just extract what the definition of $\lim \dfrac{a_n}{b_n}$ means. It says for every $\epsilon \gt 0$ there is $ n_0$ such that the difference between $ \dfrac{a_n}{b_n} $ and the limit ($ = 1$ ) is less than $ \epsilon$. So all you need to write in your paper would be to let $\epsilon $ be arbitrary. Then since $ \lim \dfrac{a_n}{b_n}$ is given, choose $ \dfrac{\epsilon}{M} $ as your arbitrary positive quantity. Then as per the definition of the limit there exists $ n_0 \in \Bbb N $ such that $$ n \ge n_0 \implies \left|{\dfrac{b_n}{a_n} - 1}\right| = \dfrac{|(b_n - a_n) - 0|}{|a_n|} \lt \dfrac{\epsilon}{M} \implies | (b_n - a_n ) - 0 | \lt |a_n | \cdot \dfrac{\epsilon}{M} \le \epsilon $$

And you're done!

$\endgroup$
  • $\begingroup$ This is the problem. i don't know what to write after i choose arbitrary epsilon. i need to someone write the answer like in a test. it's not the question of HOW to do this, its about WHAT to write to do it formally, tnx! $\endgroup$ – user2637293 Sep 18 '14 at 1:15
  • $\begingroup$ @user2637293: I edited the answer. Comment if you need more help. $\endgroup$ – Ishfaaq Sep 18 '14 at 6:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.