3
$\begingroup$

Prove that sequence {(2n+1)/n} is Cauchy.

I understand the definition of a Cauchy sequence; however, I'm not sure how to find the necessary value of N to satisfy the prove.

I know that you can simply proof that the sequence is Cauchy by stating that it converges to 2. But, for this specific problem we are asked to use strictly the definition of a Cauchy sequence in writing the proof.

Thanks for the help in advance.

$\endgroup$
2
  • 1
    $\begingroup$ Where do you get stuck? I suggest you write out what you can do, and we'll see if we can help you afterwards $\endgroup$ – davidlowryduda♦ Sep 4 '14 at 6:47
  • 1
    $\begingroup$ The proof is easier if you write $\frac{2n+1}{n} = 2+\frac{1}{n}$. $\endgroup$ – JimmyK4542 Sep 4 '14 at 6:49
4
$\begingroup$

Hint: Without loss of generality, suppose $m>n$. Then $$\Bigg\vert \frac{2n+1}{n}- \frac{2m+1}{m}\Bigg\vert= \Bigg\vert \frac{m-n}{mn}\Bigg\vert\le\Bigg\vert \frac{m}{mn}\Bigg\vert=\frac{1}{n}$$

$\endgroup$
7
  • $\begingroup$ Thank you. So N has to be greater than or equal to 1/n? Sorry that I'm unfamiliar with the formatting. $\endgroup$ – jlang Sep 4 '14 at 7:00
  • 1
    $\begingroup$ $N$ is always greater than or equal to $1/n$ if $N\in \mathbb{N}$. Your choice of $N$ should depend on a given $\varepsilon>0$. $\endgroup$ – The Substitute Sep 4 '14 at 7:04
  • $\begingroup$ Oops, I mean 1/n should be less than epsilon. For some reason I just can't grasp how to select what integer N should be based on any given epsilon. $\endgroup$ – jlang Sep 4 '14 at 7:26
  • $\begingroup$ How big should $n$ be if $1/n$ is to be less than $\varepsilon$? You want $N$ large enough so that when $n\ge N$, we have $1/n<\varepsilon$. $\endgroup$ – The Substitute Sep 4 '14 at 7:34
  • $\begingroup$ Oh, so you want N < 1/ε? $\endgroup$ – jlang Sep 4 '14 at 7:40
3
$\begingroup$

There is a theorem"a sequence is convergent iff it is a Cauchy sequence" so it needs to prove that the given sequence is convergent which emphasises that it is Cauchy.$$a_n=2+\frac{1}{n}\to 2$$i.e. the sequence is convergent and hence it is Cauchy sequence

$\endgroup$
2
  • $\begingroup$ Did you want to write that it converges to 2 (instead of 0)? $\endgroup$ – Martin Sleziak Sep 5 '14 at 14:42
  • $\begingroup$ yes yes it converges to 2,extremely sorry for the mistake $\endgroup$ – sapta Sep 6 '14 at 10:23
1
$\begingroup$

Assuming that you're considering this sequence as a sequence in $\mathbb{R}$, then the fact that $\{ \frac{2n+1}{n}\}_{n=1}^\infty$ is a convergent sequence (it converges to $2$) and using the fact that any convergent sequence is necessarily Cauchy.

$\endgroup$
3
  • $\begingroup$ I know this is probably the easiest way to proof that this sequence is Cauchy; however, for the sake of this problem we are asked strictly to use the definition of a Cauchy sequence in this proof. $\endgroup$ – jlang Sep 4 '14 at 6:50
  • $\begingroup$ You can use the proof for showing that a convergent sequence is a cauchy sequence to turn the above into such a proof. Let me know if you need some help with this $\endgroup$ – Hayden Sep 4 '14 at 6:51
  • $\begingroup$ Again, I know. However, we are asked to use the most basic definition. Namely, that a sequence is Cauchy if and only if for each epsilon greater than zero there is a positive integer N that if m, n are greater than or equal to N, then |a_n - a_m| < epsilon. We are suppose to refrain from using other criteria that would constitute a sequence as Cauchy. $\endgroup$ – jlang Sep 4 '14 at 6:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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