Proof that a given value is NOT the limit of a sequence.

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

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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
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
Have you ever proved that the limit of a sequence is unique? – Daniel Freedman Dec 9 '11 at 20:00

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

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

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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.

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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 $\frac{1}{2}$.

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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.

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