July 3rd, 2017, 07:25 AM  #1 
Senior Member Joined: Dec 2015 From: Earth Posts: 153 Thanks: 21  Limit consequence
Is consequence true ? If $\displaystyle \lim_{n\rightarrow \infty} \frac{a_n}{b_n}=0 \; \; \Rightarrow \; \; b_n\geq a_n$ 
July 3rd, 2017, 08:17 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,854 Thanks: 2228 Math Focus: Mainly analysis and algebra 
No. $a_n = 0, b_n = 1$ If you meant to have both positive \begin{align*} a_n &= n \\ b_n &= \begin{cases} n^2 & (n \gt N) \\ 1 & (n \le N) \end{cases} \end{align*} The answer to the question you intended to ask (i.e. for sufficiently large $n$) is yes. Think about $\delta\epsilon$ definition. $\displaystyle \lim_{n \to \infty} \frac{a_n}{b_n} = 0$ means that for every $\epsilon \gt 0$ there exists an $N \in \mathbb N$ such that $\frac{a_n}{b_n} \lt \epsilon$ for all $n \gt N$. Pick $ \epsilon = 1$ and you have your result. Last edited by v8archie; July 3rd, 2017 at 08:40 AM. 
July 3rd, 2017, 08:39 AM  #3 
Senior Member Joined: Dec 2015 From: Earth Posts: 153 Thanks: 21 
i forgot to check $\displaystyle a_n,b_n >0$

July 3rd, 2017, 08:40 AM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,854 Thanks: 2228 Math Focus: Mainly analysis and algebra 
See update.

July 4th, 2017, 06:40 AM  #5  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,079 Thanks: 87  Quote:
$\displaystyle \left  \frac{a_{n}}{b_{n}} \right \leq \epsilon\\ a_{n} \leqb_n$, n>N Ref: v8archie post#2. Last edited by zylo; July 4th, 2017 at 06:51 AM. Reason: add n>N  

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