Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Question: prove that the sequences $\left\{\frac{n}{2n+3}\right\}$ and $\left\{\frac{n}{2n-3}\right\}$ converge using the definition.

What I have: I know both of them have limit $1/2$.

The definition says: a sequence $\{x_n\}$ is said to converge to a number $x \in \mathbb R$, if for every $\epsilon >0$, there exists an $M \in \mathbb N$ such that $|x_n-x|<\epsilon$ for all $n\geq M$.

Start: let $\epsilon >0$ such that.... Where do I go from here?

share|improve this question
    
Then you need to find $M$ such that .... –  John Nov 5 '13 at 7:32
    
Let e>0 such that M element N and 1/M+1<e. then for any n>=M ...is this correct thus far? –  Cynthia Nov 5 '13 at 7:36
    
The goal is to get $|\frac{n}{2n+3} -\frac{1}{2}| < \epsilon$. You might play around with this inequality to found how large should $n$ be. –  John Nov 5 '13 at 7:37

2 Answers 2

up vote 2 down vote accepted

Let $\varepsilon >0$. We want do show the existence of a number $M\in \mathbb{N}$ such that $\left|\frac{n}{2n+3}-\frac{1}{2}\right|< \varepsilon$ if $n\ge M$.

$$\left|\frac{n}{2n+3}-\frac{1}{2}\right|=\frac{3}{2(2n+3)}<\varepsilon\iff \frac{1}{2}\left( \frac{3}{2\varepsilon}-3\right)<n$$ so if we choose $M> \frac{1}{2}\left( \frac{3}{2\varepsilon}-3\right),$ we get $\left|\frac{n}{2n+3}-\frac{1}{2}\right|< \varepsilon$ if $n\ge M$.

share|improve this answer
    
Would it work the same way for sequence {n/(2*n-3)}? Thank you. –  Cynthia Nov 5 '13 at 7:48
    
@Cynthia Yes, the argument is essentially the same. –  Ron Ford Nov 5 '13 at 8:03

$\frac{n}{2n+3} = \frac{1}{2+3/n}$ so as n goes to infinity $3/n$ approaches zero. So the limit of the sequence is $\frac{1}{2}$

share|improve this answer
2  
I think she might want a $\epsilon$-$N$ approach. Your answer is absolutely correct. –  John Nov 5 '13 at 7:36
    
Yea that was my mistake. I just caught that. –  RDizzl3 Nov 5 '13 at 7:38

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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