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.

In the sequence:

$$\lim _{n\rightarrow \infty }{\frac {n+1}{2\,n+3}}\neq 3$$

I know how to prove that the limit is actually $1/2$, but is there another way to prove that 1 is NOT the limit?

I tried to prove by negation showing that if 1 is the limit I can't find an $N$ that for every epsilon etc etc but got a bit lost. I know this is trivial but would appreciate the help. Thanks

share|improve this question
    
i tried copying this in as Latex. Don't really know what i'm doing. –  nofe Dec 9 '11 at 19:55
    
Hi nofeliram, welcome to the site! To get the LaTeX to work, you have to put dollar signs around it; see the quick intro here. –  Zev Chonoles Dec 9 '11 at 19:57
1  
To prove that $1$ is not the limit, you have to find an $\epsilon$ such that for all $N$ there exists an $n>N$ such that $|1-\frac{n+1}{2n+3}|\geq \epsilon$. In this case, it's fairly easy, because, for $n>0$, $\frac{n+1}{2n+3}$ is always less then $\frac{1}{2}$, so $\epsilon=\frac{1}{2}$, for all $n>0$, $|1-\frac{n+1}{2n+3}|>\epsilon$. –  Thomas Andrews Dec 9 '11 at 20:00
2  
Have you ever proved that the limit of a sequence is unique? –  Daniel Freedman Dec 9 '11 at 20:00

4 Answers 4

up vote 9 down vote accepted

For example, show the limit is $\le 1/2$ like this:

$$ \frac{n+1}{2n+3} < \frac{n+1}{2n+2} = \frac{1}{2} $$

share|improve this answer

Yes, the value of the fraction is always below 1/2. So it stays far away from 3, indeed never gets within 5/2 of it. Note that it is not necessary to know that there is any limit to do answer your question. Anyway, take $\varepsilon = 2$ and no $n$ ever gets you within $\varepsilon$ of 3.

share|improve this answer

Let $X$ be some number other than $\frac{1}{2}$. Let's define $d=|X-\frac{1}{2}|$, and because $X\neq\frac{1}{2}$ we have $d>0$.

The reverse triangle inequality says that for any $a$ and $b$, $$|a-b|\geq ||a|-|b||.$$ In particular, $$\left|X-\frac{n+1}{2n+3}\right|=\left|X-\left(\frac{n+\frac{3}{2}}{2n+3}-\frac{\frac{1}{2}}{2n+3}\right)\right|=\left|\left(X-\frac{1}{2}\right)-\left(-\frac{1}{4n+6}\right)\right|\geq d-\frac{1}{4n+6}$$

Suppose that $$\lim_{n\to\infty}\frac{n+1}{2n+3}=X,$$ i.e. for any $\epsilon>0$, there exists an $N\in\mathbb{N}$ such that: for all $n>N$, $$\left|X-\frac{n+1}{2n+3}\right|<\epsilon.$$ Then we have that for all $n>N$, $$d-\frac{1}{4n+6}<\epsilon,$$ or equivalently$$d<\epsilon+\frac{1}{4n+6}.$$ But this is true for all $n>N$ iff $d\leq \epsilon$. And $d\leq \epsilon$ for all $\epsilon>0$ iff $d=0$. But $d>0$; contradiction.

Thus, the limit cannot be anything other than $\frac{1}{2}$.

share|improve this answer

Let $\epsilon=1/4$. Let $N$ be given. Then $${n+1\over 2n+3}\le {n \over 2n}\le{1\over2}\quad\Rightarrow|{n+1\over 2n+3} -1|\ge 1/2.$$ This is true for all $n\ge N$.

So, there is no $N$ such that $|{n+1\over 2n+3} - 1|<\epsilon$ for all $n\ge N$. This shows the sequence does not converge to 1.

share|improve this answer

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.