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?


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

  • $\begingroup$ 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


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!

  • $\begingroup$ 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

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