Proving that the sequence $\{\frac{3n+5}{2n+6}\}$ is Cauchy.

Question

I'm not quite sure how to tackle these kinds of questions in general, but I tried something that I thought could be right. Hoping to be steered in the right direction here!

Let $\{\frac{3n+5}{2n+6}\}$ be a sequence of real numbers. Prove that this sequence is Cauchy.

Proof:

We want to establish that $\forall_{\epsilon>0}\exists_{{n_0}\in{\mathbb{N}}}\forall_{n,m\geq n_0}\big(|f(n)-f(m)|\big)<\epsilon.$

Suppose $n>m$ without loss of generality. We then know that $\frac{3n+5}{2n+6}>\frac{3m+5}{2m+6}$ and thus that $\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}>0$ such that $\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}=|\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}|=|f(n)-f(m)|.$

Let us work out the original sequence:

$\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}=\frac{(3n+5)(2m+6)-(3m+5)(2n+6)}{(2n+6)(2m+6)} = \frac{8(n-m)}{(n2+6)(2m+6)}<\frac{8(n-m)}{nm}= 8(\frac{1}{n}- \frac{1}{m}).$

We know that $\frac{1}{n}<\frac{1}{m}$ as $n>m$ and that $\frac{1}{n}\leq\frac{1}{n_0}, \frac{1}{m}\leq\frac{1}{n_0}$ for $n,m\geq n_0$.

This means that $\frac{1}{n}-\frac{1}{m}\leq \frac{1}{n_0}- \frac{1}{m}\leq\frac{1}{n_0}$, and thus $8(\frac{1}{n}- \frac{1}{m})\leq \frac{8}{n_0}$.

Let $\epsilon=\frac{8}{n_0}$, as it only depends on $n_0$ it can become arbitrarily small.

Then the following inequality holds: $\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}<8(\frac{1}{n}- \frac{1}{m})\leq \frac{8}{n_0}$.

So: $\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}<\epsilon$, and thus the sequence is Cauchy.$\tag*{$\Box$}$

Answer 1

$|\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}|=\frac{|n-m|}{(2n+6)(2m+6)}\le\frac{8|n-m|}{nm} =8|\frac{1}{n}-\frac{1}{m}|$.

Can you proceed ?

Answer 2

As you were told in the comments, you must take some $\varepsilon>0$ and then prove that $\left|\frac{3n+5}{2m+6}-\frac{3n+5}{2n+6}\right|<\varepsilon$ if $m$ and $n$ are large enough.

Note that$$(\forall n\in\mathbb{N}):\frac{3n+5}{2n+6}=\frac{3n+9}{2n+6}-\frac4{2n+6}=\frac32-\frac2{n+3}.$$So, take $p\in\mathbb N$ such that $\frac2{p+3}<\frac\varepsilon2$. Then$$m,n\geqslant p\implies\left|\frac{3m+5}{2m+6}-\frac{3n+5}{2n+6}\right|=\left|\left(\frac32-\frac2{m+3}\right)-\left(\frac32-\frac2{n+3}\right)\right|<\varepsilon.$$

Answer 3

If you do have to prove it with the ε-$n_0$ method, it's much simpler that that:

First rewrite $a_n$ as $$a_n=\frac{3n+5}{2n+6}=\frac32-\frac2{n+3}.$$ Then, if $m,n>n_0$, $$\bigl|a_m-a_n\bigr|=2\,\biggl|\frac1{m+3}-\frac1{n+3}\biggr|=\frac{ 2\,| m-n|}{(m+3)(n+3)}\le 2\biggl|\frac1m-\frac1n\biggr|<2\biggl(\frac1m+\frac1n\biggr)<\frac4{n_0},$$ and this one is less than $\varepsilon$ as soon as $\;n_0>\dfrac 4\varepsilon$.

Answer 4

Your proof doesn't use any properties of the sequence: you saying that the sequence is Cauchy is no proof. Also, your first sentence is wrong, as the sequence is increasing.

What you want to do is $$ \frac {3n+5}{2n+6}-\frac {3m+5}{2m+6}=\frac{8 (n-m)}{(2m+6)(2n+6)} \leq \frac{8 (n-m)}{4mn}=\frac2m-\frac2n. $$ Now you can show that, the bigger $n,m $ are, the smaller the difference. Usually one needs absolute value, but in this particular case the condition $n>m $ guarantees that everything above is positive.

Answer 5

Let $ \epsilon >0$ be an arbitrary small number.

Choose a positive integer $n_0$ such that $$\frac {2}{n_0 +3} <\frac {\epsilon }{2}$$

Note that if $m \ge n_0$, and $n\ge n_0$ are integers, then $$ |f(n)-f(m)| = |\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}|= $$ $$|\frac {2(m-n)}{(n+3)(m+3)} |\le $$ $$\frac {2(m+n)}{(n+3)(m+3)} <$$ $$\frac {2}{(m+3)}+\frac {2}{(n+3)} <\epsilon $$

Thus the sequence ($ \frac{3n+5}{2n+6}$) is a Cauchy sequence.