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Show $x_n = sqrtn$ is not Cauchy sequence

Cauchy sequences. Show that $(x_n)$ is Cauchy.Prove that the sequence $(f(x_n))_ngeqslant1$ is Cauchy.Is being a Cauchy sequence equivalent to $ lim_nto+inftyd(x_n+k,x_n)=0$ for every $k$?Show that the sequence $a_n=1+frac1sqrt2 +frac1sqrt3 + dots + frac1sqrtn$ is not CauchyIf $x_n:=sqrtn$, show that $(x_n)$ satisfies $lim|x_n+1-x_n|=0$, but that it is not a Cauchy sequence.Cauchy Sequence in Normed SpaceHow to prove that is a Cauchy sequenceDefinition of Cauchy SequenceProve that a Cauchy sequence is convergentCauchy sequences always have a largest or smallest element past an arbitrary index

0

$begingroup$

Consider the sequence $x_n$,$x_n$=$sqrtn$

Show that $forall varepsilon > 0, exists n_0 in Bbb N$ s.t. $forall n geq n_0$, |$x_n+1-x_n$|<$varepsilon$.

This is what I have:

Let $varepsilon>0$ and let $n_0$ = $(frac12varepsilon)^2, forall n geq n_0$. $|sqrtn+1-sqrtn|$. So, $|frac(sqrtn+1-sqrtn)(sqrtn+1+sqrtn)(sqrtn+1+sqrtn)|$.

Then, $|frac1sqrtn+1$ + $sqrtn|$ $leq$ $|frac1sqrtn+sqrtn|$=$|frac12sqrtn|$= $varepsilon$.

Therefore, $|x_n+1-x_n|<varepsilon, forall n geq n_0$

After I did all this it got me thinking can you prove $x_n$ is not cauchy? if so how?

real-analysis analysis cauchy-sequences

share|cite|improve this question

edited Mar 11 at 19:45

Andrews

1,2691421

asked Mar 11 at 19:22

user597188user597188

234

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You made a typo, $|frac1sqrtn+1 + sqrtn|$ probably needs to be $|frac1sqrtn+1 + sqrtn|$. Your proof is correct. Note that for Cauchy you need that for any $m,ngeq n_0$: $|x_n-x_m|<epsilon$ which is not the case
  $endgroup$
  – Stan Tendijck
  Mar 11 at 19:32
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Consider the sequence $x_n$,$x_n$=$sqrtn$

Show that $forall varepsilon > 0, exists n_0 in Bbb N$ s.t. $forall n geq n_0$, |$x_n+1-x_n$|<$varepsilon$.

This is what I have:

Let $varepsilon>0$ and let $n_0$ = $(frac12varepsilon)^2, forall n geq n_0$. $|sqrtn+1-sqrtn|$. So, $|frac(sqrtn+1-sqrtn)(sqrtn+1+sqrtn)(sqrtn+1+sqrtn)|$.

Then, $|frac1sqrtn+1$ + $sqrtn|$ $leq$ $|frac1sqrtn+sqrtn|$=$|frac12sqrtn|$= $varepsilon$.

Therefore, $|x_n+1-x_n|<varepsilon, forall n geq n_0$

After I did all this it got me thinking can you prove $x_n$ is not cauchy? if so how?

real-analysis analysis cauchy-sequences

share|cite|improve this question

edited Mar 11 at 19:45

Andrews

1,2691421

asked Mar 11 at 19:22

user597188user597188

234

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You made a typo, $|frac1sqrtn+1 + sqrtn|$ probably needs to be $|frac1sqrtn+1 + sqrtn|$. Your proof is correct. Note that for Cauchy you need that for any $m,ngeq n_0$: $|x_n-x_m|<epsilon$ which is not the case
  $endgroup$
  – Stan Tendijck
  Mar 11 at 19:32
  
  

  
  
  
  

add a comment | 

$begingroup$

Consider the sequence $x_n$,$x_n$=$sqrtn$

Show that $forall varepsilon > 0, exists n_0 in Bbb N$ s.t. $forall n geq n_0$, |$x_n+1-x_n$|<$varepsilon$.

This is what I have:

Let $varepsilon>0$ and let $n_0$ = $(frac12varepsilon)^2, forall n geq n_0$. $|sqrtn+1-sqrtn|$. So, $|frac(sqrtn+1-sqrtn)(sqrtn+1+sqrtn)(sqrtn+1+sqrtn)|$.

Then, $|frac1sqrtn+1$ + $sqrtn|$ $leq$ $|frac1sqrtn+sqrtn|$=$|frac12sqrtn|$= $varepsilon$.

Therefore, $|x_n+1-x_n|<varepsilon, forall n geq n_0$

After I did all this it got me thinking can you prove $x_n$ is not cauchy? if so how?

real-analysis analysis cauchy-sequences

share|cite|improve this question

edited Mar 11 at 19:45

Andrews

1,2691421

asked Mar 11 at 19:22

user597188user597188

234

$endgroup$

share|cite|improve this question

edited Mar 11 at 19:45

Andrews

1,2691421

asked Mar 11 at 19:22

user597188user597188

234

share|cite|improve this question

edited Mar 11 at 19:45

Andrews

1,2691421

asked Mar 11 at 19:22

user597188user597188

234

edited Mar 11 at 19:45

Andrews

1,2691421

edited Mar 11 at 19:45

Andrews

1,2691421

asked Mar 11 at 19:22

user597188user597188

234

asked Mar 11 at 19:22

user597188user597188

234

3 Answers
3

active

oldest

votes

1

$begingroup$

Any Cauchy sequence converges. Since $sqrtnto+infty$, it diverges, so it is not Cauchy.

Telling about the condition in question, we have $$sqrtn+1-sqrtn=frac1sqrtn+1+sqrtnto 0,$$ so it holds trivially by definition of a limit.

share|cite|improve this answer

answered Mar 11 at 19:54

szw1710szw1710

6,4801223

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You forgot "in a complete metric space" after "any Cauchy sequence converges." I am guessing that they have not proven that $mathbbR$ is complete and they will have to prove the non-Cauchy property "by hand".
  $endgroup$
  – Jair Taylor
  Mar 11 at 19:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JairTaylor, If the metric os not specified, usually one thinks about the Eucldean one, which is, of course, complete. In OP's question we have $|x_n+1-x_n|$, which clearly indicates the Euclidean metric. Please show me, where OP specifies another metric.
  $endgroup$
  – szw1710
  Mar 11 at 23:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Of course, it is implied in this case the metric is Euclidean. But it is important to note that Cauchy sequences do not necessarily converge in other metrics.
  $endgroup$
  – Jair Taylor
  Mar 11 at 23:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  More importantly, this answer relies on the completeness of (and construction of) $mathbbR$ which is not trivial. But of course it is correct.
  $endgroup$
  – Jair Taylor
  Mar 11 at 23:47
  
  

  
  
  
  

add a comment | 

1

$begingroup$

You have a solid argument for the main part, aside from the typo $|frac1sqrtn+1+sqrtn|$ in the second line (already noted in a comment).

After I did all this it got me thinking can you prove $x_n$ is not Cauchy? If so, how?

The definition of a Cauchy sequence refers not only to the difference between adjacent pairs of elements, but also the difference between pairs of elements any index distance apart - as long as they're both far enough out in the sequence.

Just eyeballing it, the square roots increase without bound; we should be able to find one that exceeds any particular target. So, then, given some $m$, can you find some larger $n$ with $sqrtn-sqrtm > 1$? A rule to do this would immediately contradict the Cauchy criterion for $epsilon=1$, and show the sequence isn't Cauchy.

Don't worry about being terribly precise here. We don't need the smallest $n$, just anything that works.

share|cite|improve this answer

answered Mar 11 at 19:56

jmerryjmerry

14.2k1629

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  so if I chose n to be 1/4 or 1/8 would either of those work?
  $endgroup$
  – user597188
  Mar 11 at 20:36
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Uh, what? $n$ is an integer. And, for the anti-Cauchy criterion we're trying to prove, it has to be greater than $m$.
  $endgroup$
  – jmerry
  Mar 11 at 21:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  sorry I meant like n/2$sqrtm$<1/4 or 1/8
  $endgroup$
  – user597188
  Mar 11 at 21:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I still doubt that's what you mean; there's no reason to divide $n$ and $m$. Now, in what I wrote, the $1$ is arbitrary. We could replace it by $1/4$, or by $40$. But, whatever we do, it has to be a constant, not depending on $m$ at all.
  $endgroup$
  – jmerry
  Mar 11 at 21:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So if I let $epsilon$ =1/4 id have (m-n)/( $sqrtm$ + $sqrtn$) $geq$ n/(2$sqrtm$) then $sqrtm$/2 - n/(2$sqrtm$) > 1/2-n/(2$sqrtm$) >1/2-1/4 > 1/4 which is what epsilon equals but how would I do it if I let epsilon=1?
  $endgroup$
  – user597188
  Mar 11 at 21:18
  
  

  
  
  
  

 | 
show 2 more comments

1

$begingroup$

Let $epsilon_0=frac13$. For all $ninmathbbN$ and $p=n$,
$$ |sqrtn+p-sqrt n|=(sqrt2-1)sqrt ngesqrt 2-1>epsilon_0. $$
Namely, $sqrtn$ is not Cauchy.

share|cite|improve this answer

answered Mar 12 at 14:44

xpaulxpaul

23.3k24655

$endgroup$

add a comment | 

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1

$begingroup$

Any Cauchy sequence converges. Since $sqrtnto+infty$, it diverges, so it is not Cauchy.

Telling about the condition in question, we have $$sqrtn+1-sqrtn=frac1sqrtn+1+sqrtnto 0,$$ so it holds trivially by definition of a limit.

share|cite|improve this answer

answered Mar 11 at 19:54

szw1710szw1710

6,4801223

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You forgot "in a complete metric space" after "any Cauchy sequence converges." I am guessing that they have not proven that $mathbbR$ is complete and they will have to prove the non-Cauchy property "by hand".
  $endgroup$
  – Jair Taylor
  Mar 11 at 19:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JairTaylor, If the metric os not specified, usually one thinks about the Eucldean one, which is, of course, complete. In OP's question we have $|x_n+1-x_n|$, which clearly indicates the Euclidean metric. Please show me, where OP specifies another metric.
  $endgroup$
  – szw1710
  Mar 11 at 23:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Of course, it is implied in this case the metric is Euclidean. But it is important to note that Cauchy sequences do not necessarily converge in other metrics.
  $endgroup$
  – Jair Taylor
  Mar 11 at 23:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  More importantly, this answer relies on the completeness of (and construction of) $mathbbR$ which is not trivial. But of course it is correct.
  $endgroup$
  – Jair Taylor
  Mar 11 at 23:47
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Any Cauchy sequence converges. Since $sqrtnto+infty$, it diverges, so it is not Cauchy.

Telling about the condition in question, we have $$sqrtn+1-sqrtn=frac1sqrtn+1+sqrtnto 0,$$ so it holds trivially by definition of a limit.

share|cite|improve this answer

answered Mar 11 at 19:54

szw1710szw1710

6,4801223

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You forgot "in a complete metric space" after "any Cauchy sequence converges." I am guessing that they have not proven that $mathbbR$ is complete and they will have to prove the non-Cauchy property "by hand".
  $endgroup$
  – Jair Taylor
  Mar 11 at 19:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JairTaylor, If the metric os not specified, usually one thinks about the Eucldean one, which is, of course, complete. In OP's question we have $|x_n+1-x_n|$, which clearly indicates the Euclidean metric. Please show me, where OP specifies another metric.
  $endgroup$
  – szw1710
  Mar 11 at 23:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Of course, it is implied in this case the metric is Euclidean. But it is important to note that Cauchy sequences do not necessarily converge in other metrics.
  $endgroup$
  – Jair Taylor
  Mar 11 at 23:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  More importantly, this answer relies on the completeness of (and construction of) $mathbbR$ which is not trivial. But of course it is correct.
  $endgroup$
  – Jair Taylor
  Mar 11 at 23:47
  
  

  
  
  
  

add a comment | 

$begingroup$

Any Cauchy sequence converges. Since $sqrtnto+infty$, it diverges, so it is not Cauchy.

Telling about the condition in question, we have $$sqrtn+1-sqrtn=frac1sqrtn+1+sqrtnto 0,$$ so it holds trivially by definition of a limit.

share|cite|improve this answer

answered Mar 11 at 19:54

szw1710szw1710

6,4801223

$endgroup$

share|cite|improve this answer

answered Mar 11 at 19:54

szw1710szw1710

6,4801223

answered Mar 11 at 19:54

szw1710szw1710

6,4801223

answered Mar 11 at 19:54

szw1710szw1710

6,4801223

1

$begingroup$

You have a solid argument for the main part, aside from the typo $|frac1sqrtn+1+sqrtn|$ in the second line (already noted in a comment).

After I did all this it got me thinking can you prove $x_n$ is not Cauchy? If so, how?

The definition of a Cauchy sequence refers not only to the difference between adjacent pairs of elements, but also the difference between pairs of elements any index distance apart - as long as they're both far enough out in the sequence.

Just eyeballing it, the square roots increase without bound; we should be able to find one that exceeds any particular target. So, then, given some $m$, can you find some larger $n$ with $sqrtn-sqrtm > 1$? A rule to do this would immediately contradict the Cauchy criterion for $epsilon=1$, and show the sequence isn't Cauchy.

Don't worry about being terribly precise here. We don't need the smallest $n$, just anything that works.

share|cite|improve this answer

answered Mar 11 at 19:56

jmerryjmerry

14.2k1629

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  so if I chose n to be 1/4 or 1/8 would either of those work?
  $endgroup$
  – user597188
  Mar 11 at 20:36
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Uh, what? $n$ is an integer. And, for the anti-Cauchy criterion we're trying to prove, it has to be greater than $m$.
  $endgroup$
  – jmerry
  Mar 11 at 21:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  sorry I meant like n/2$sqrtm$<1/4 or 1/8
  $endgroup$
  – user597188
  Mar 11 at 21:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I still doubt that's what you mean; there's no reason to divide $n$ and $m$. Now, in what I wrote, the $1$ is arbitrary. We could replace it by $1/4$, or by $40$. But, whatever we do, it has to be a constant, not depending on $m$ at all.
  $endgroup$
  – jmerry
  Mar 11 at 21:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So if I let $epsilon$ =1/4 id have (m-n)/( $sqrtm$ + $sqrtn$) $geq$ n/(2$sqrtm$) then $sqrtm$/2 - n/(2$sqrtm$) > 1/2-n/(2$sqrtm$) >1/2-1/4 > 1/4 which is what epsilon equals but how would I do it if I let epsilon=1?
  $endgroup$
  – user597188
  Mar 11 at 21:18
  
  

  
  
  
  

 | 
show 2 more comments

1

$begingroup$

You have a solid argument for the main part, aside from the typo $|frac1sqrtn+1+sqrtn|$ in the second line (already noted in a comment).

After I did all this it got me thinking can you prove $x_n$ is not Cauchy? If so, how?

The definition of a Cauchy sequence refers not only to the difference between adjacent pairs of elements, but also the difference between pairs of elements any index distance apart - as long as they're both far enough out in the sequence.

Just eyeballing it, the square roots increase without bound; we should be able to find one that exceeds any particular target. So, then, given some $m$, can you find some larger $n$ with $sqrtn-sqrtm > 1$? A rule to do this would immediately contradict the Cauchy criterion for $epsilon=1$, and show the sequence isn't Cauchy.

Don't worry about being terribly precise here. We don't need the smallest $n$, just anything that works.

share|cite|improve this answer

answered Mar 11 at 19:56

jmerryjmerry

14.2k1629

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  so if I chose n to be 1/4 or 1/8 would either of those work?
  $endgroup$
  – user597188
  Mar 11 at 20:36
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Uh, what? $n$ is an integer. And, for the anti-Cauchy criterion we're trying to prove, it has to be greater than $m$.
  $endgroup$
  – jmerry
  Mar 11 at 21:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  sorry I meant like n/2$sqrtm$<1/4 or 1/8
  $endgroup$
  – user597188
  Mar 11 at 21:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I still doubt that's what you mean; there's no reason to divide $n$ and $m$. Now, in what I wrote, the $1$ is arbitrary. We could replace it by $1/4$, or by $40$. But, whatever we do, it has to be a constant, not depending on $m$ at all.
  $endgroup$
  – jmerry
  Mar 11 at 21:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So if I let $epsilon$ =1/4 id have (m-n)/( $sqrtm$ + $sqrtn$) $geq$ n/(2$sqrtm$) then $sqrtm$/2 - n/(2$sqrtm$) > 1/2-n/(2$sqrtm$) >1/2-1/4 > 1/4 which is what epsilon equals but how would I do it if I let epsilon=1?
  $endgroup$
  – user597188
  Mar 11 at 21:18
  
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

You have a solid argument for the main part, aside from the typo $|frac1sqrtn+1+sqrtn|$ in the second line (already noted in a comment).

After I did all this it got me thinking can you prove $x_n$ is not Cauchy? If so, how?

The definition of a Cauchy sequence refers not only to the difference between adjacent pairs of elements, but also the difference between pairs of elements any index distance apart - as long as they're both far enough out in the sequence.

Just eyeballing it, the square roots increase without bound; we should be able to find one that exceeds any particular target. So, then, given some $m$, can you find some larger $n$ with $sqrtn-sqrtm > 1$? A rule to do this would immediately contradict the Cauchy criterion for $epsilon=1$, and show the sequence isn't Cauchy.

Don't worry about being terribly precise here. We don't need the smallest $n$, just anything that works.

share|cite|improve this answer

answered Mar 11 at 19:56

jmerryjmerry

14.2k1629

$endgroup$

share|cite|improve this answer

answered Mar 11 at 19:56

jmerryjmerry

14.2k1629

answered Mar 11 at 19:56

jmerryjmerry

14.2k1629

answered Mar 11 at 19:56

jmerryjmerry

14.2k1629

1

$begingroup$

Let $epsilon_0=frac13$. For all $ninmathbbN$ and $p=n$,
$$ |sqrtn+p-sqrt n|=(sqrt2-1)sqrt ngesqrt 2-1>epsilon_0. $$
Namely, $sqrtn$ is not Cauchy.

share|cite|improve this answer

answered Mar 12 at 14:44

xpaulxpaul

23.3k24655

$endgroup$

add a comment | 

1

$begingroup$

Let $epsilon_0=frac13$. For all $ninmathbbN$ and $p=n$,
$$ |sqrtn+p-sqrt n|=(sqrt2-1)sqrt ngesqrt 2-1>epsilon_0. $$
Namely, $sqrtn$ is not Cauchy.

share|cite|improve this answer

answered Mar 12 at 14:44

xpaulxpaul

23.3k24655

$endgroup$

add a comment | 

$begingroup$

Let $epsilon_0=frac13$. For all $ninmathbbN$ and $p=n$,
$$ |sqrtn+p-sqrt n|=(sqrt2-1)sqrt ngesqrt 2-1>epsilon_0. $$
Namely, $sqrtn$ is not Cauchy.

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answered Mar 12 at 14:44

xpaulxpaul

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$endgroup$

share|cite|improve this answer

answered Mar 12 at 14:44

xpaulxpaul

23.3k24655

answered Mar 12 at 14:44

xpaulxpaul

23.3k24655

answered Mar 12 at 14:44

xpaulxpaul

23.3k24655