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$(a_n)_n=1^infty$ is a convergent sequence of integers. Prove the existence of $N in mathbbN$ such that $a_i = a_j$ for all $i, j > N$. [duplicate]

Let $(a_n)_n$ be a convergent sequence of integers , what can we say about $(a_n)_n$If $x_n$ is a monotone sequence and there exists $k$ such that $lim_n rightarrow infty x_n = x_k$, then $x_n=x_k$ for $n>k$Prove there is a sequence of increasing positive integers $n_i$ s.t. the limit of $lim_n_i to infty sin(n_i)$ existsProof of the Existence of the Scalar Multiple of a Convergent SequenceShow that $ lim_n rightarrow infty fracn!2^n = infty $Prove that if $(a_n)$ is a monotone increasing sequence of integers then $lim_ntoinfty(1+frac1a_n)^a_n=e$$(a_n)$ is a monotone increasing sequence of integers. Prove that: $lim_ntoinfty(1+frac1a_n)^a_n=e$Convergent sequences of integers are eventually constant.How to prove that, if a sequence of integers converges, then its limit is an integer?$forallepsilon>0, exists NinBbb N$ such that $forall n>mge N, a_n-a_m<epsilon$. Prove that $a_n$ converges to a real limit or to $-infty$If $a_n$ is a sequence such that $0leq a_m+nleq a_m+a_n$, show that $lim_ntoinftyfraca_nn$ exists

0

$begingroup$

This question already has an answer here:

• 
  Let $(a_n)_n$ be a convergent sequence of integers , what can we say about $(a_n)_n$
  
  1 answer
  
  

Assume $(a_n)_n=1^infty$ is a convergent sequence of integers. Prove
the existence of $NinmathbbN$ such that $a_i = a_j$
for all $i, j > N.$

I am completely lost on what this question is asking, do I use the definition of a limit to show it converges and doesn't leave epsilon?
I have searched and can't find this precise question.

limits

share|cite|improve this question

edited Mar 13 at 10:37

Yadati Kiran

2,1061621

asked Mar 13 at 10:19

john smithjohn smith

64

New contributor

john smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

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marked as duplicate by José Carlos Santos, Song, Dietrich Burde, Pedro Tamaroff♦ Mar 13 at 10:55

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment | 

0

$begingroup$

This question already has an answer here:

• 
  Let $(a_n)_n$ be a convergent sequence of integers , what can we say about $(a_n)_n$
  
  1 answer
  
  

Assume $(a_n)_n=1^infty$ is a convergent sequence of integers. Prove
the existence of $NinmathbbN$ such that $a_i = a_j$
for all $i, j > N.$

I am completely lost on what this question is asking, do I use the definition of a limit to show it converges and doesn't leave epsilon?
I have searched and can't find this precise question.

limits

share|cite|improve this question

edited Mar 13 at 10:37

Yadati Kiran

2,1061621

asked Mar 13 at 10:19

john smithjohn smith

64

New contributor

john smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

marked as duplicate by José Carlos Santos, Song, Dietrich Burde, Pedro Tamaroff♦ Mar 13 at 10:55

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment | 

$begingroup$

This question already has an answer here:

• 
  Let $(a_n)_n$ be a convergent sequence of integers , what can we say about $(a_n)_n$
  
  1 answer
  
  

Assume $(a_n)_n=1^infty$ is a convergent sequence of integers. Prove
the existence of $NinmathbbN$ such that $a_i = a_j$
for all $i, j > N.$

I am completely lost on what this question is asking, do I use the definition of a limit to show it converges and doesn't leave epsilon?
I have searched and can't find this precise question.

limits

share|cite|improve this question

edited Mar 13 at 10:37

Yadati Kiran

2,1061621

asked Mar 13 at 10:19

john smithjohn smith

64

New contributor

john smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Mar 13 at 10:37

Yadati Kiran

2,1061621

asked Mar 13 at 10:19

john smithjohn smith

64

New contributor

john smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Mar 13 at 10:37

Yadati Kiran

2,1061621

asked Mar 13 at 10:19

john smithjohn smith

64

New contributor

john smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Mar 13 at 10:37

Yadati Kiran

2,1061621

edited Mar 13 at 10:37

Yadati Kiran

2,1061621

asked Mar 13 at 10:19

john smithjohn smith

64

New contributor

john smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 13 at 10:19

john smithjohn smith

64

3 Answers
3

active

oldest

votes

2

$begingroup$

Let $a_n to a$. Then there exist $N$ such that $|a_n-a_m| leq |a_n-a|+|a_m-a| <frac 1 2+frac 1 2$ for all $n,m geq N$. But $a_m$ and $a_m$ are integers, so $|a_n-a_m| <1$ implies $a_n=a_m$ for all $n,m geq N$.

share|cite|improve this answer

answered Mar 13 at 10:23

Kavi Rama MurthyKavi Rama Murthy

68.1k53068

$endgroup$

add a comment | 

1

$begingroup$

If $(a_n)$ is convergent, then $(a_n)$ is Cauchy. Hence there is $N in mathbb N$ such that

$|a_n-a_m|<1$ for $m,n > N$. This gives $a_n=a_m$ for $m,n > N$, since all $a_n in mathbb Z.$

share|cite|improve this answer

answered Mar 13 at 10:25

FredFred

48.5k11849

$endgroup$

add a comment | 

0

$begingroup$

Option.

Since $a_n in mathbbZ^+$, $n=1,2,3,.....,$ is convergent in $mathbbR$ it is Cauchy.

Let $epsilon <1$ given.

There is a $N$ s.t. for $n ge mge N$

$|a_n-a_m| <1$ , since $a_n,a_m$ are integers we have

$a_n=a_m$.

share|cite|improve this answer

answered Mar 13 at 10:45

Peter SzilasPeter Szilas

11.6k2822

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  is an and am simply referring to the ai and aj in the question?
  $endgroup$
  – john smith
  Mar 13 at 10:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @johnsmith Yes, exactly.
  $endgroup$
  – Pedro Tamaroff♦
  Mar 13 at 10:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  John.Yes! My comment disappeared.Pedro answered already.n,m are dummies.You can rewrite the whole thing with i j, instead.Ok?
  $endgroup$
  – Peter Szilas
  Mar 13 at 10:58
  

  
  
  
  

add a comment | 

2

$begingroup$

Let $a_n to a$. Then there exist $N$ such that $|a_n-a_m| leq |a_n-a|+|a_m-a| <frac 1 2+frac 1 2$ for all $n,m geq N$. But $a_m$ and $a_m$ are integers, so $|a_n-a_m| <1$ implies $a_n=a_m$ for all $n,m geq N$.

share|cite|improve this answer

answered Mar 13 at 10:23

Kavi Rama MurthyKavi Rama Murthy

68.1k53068

$endgroup$

add a comment | 

2

$begingroup$

Let $a_n to a$. Then there exist $N$ such that $|a_n-a_m| leq |a_n-a|+|a_m-a| <frac 1 2+frac 1 2$ for all $n,m geq N$. But $a_m$ and $a_m$ are integers, so $|a_n-a_m| <1$ implies $a_n=a_m$ for all $n,m geq N$.

share|cite|improve this answer

answered Mar 13 at 10:23

Kavi Rama MurthyKavi Rama Murthy

68.1k53068

$endgroup$

add a comment | 

$begingroup$

Let $a_n to a$. Then there exist $N$ such that $|a_n-a_m| leq |a_n-a|+|a_m-a| <frac 1 2+frac 1 2$ for all $n,m geq N$. But $a_m$ and $a_m$ are integers, so $|a_n-a_m| <1$ implies $a_n=a_m$ for all $n,m geq N$.

share|cite|improve this answer

answered Mar 13 at 10:23

Kavi Rama MurthyKavi Rama Murthy

68.1k53068

$endgroup$

share|cite|improve this answer

answered Mar 13 at 10:23

Kavi Rama MurthyKavi Rama Murthy

68.1k53068

answered Mar 13 at 10:23

Kavi Rama MurthyKavi Rama Murthy

68.1k53068

answered Mar 13 at 10:23

Kavi Rama MurthyKavi Rama Murthy

68.1k53068

1

$begingroup$

If $(a_n)$ is convergent, then $(a_n)$ is Cauchy. Hence there is $N in mathbb N$ such that

$|a_n-a_m|<1$ for $m,n > N$. This gives $a_n=a_m$ for $m,n > N$, since all $a_n in mathbb Z.$

share|cite|improve this answer

answered Mar 13 at 10:25

FredFred

48.5k11849

$endgroup$

add a comment | 

1

$begingroup$

If $(a_n)$ is convergent, then $(a_n)$ is Cauchy. Hence there is $N in mathbb N$ such that

$|a_n-a_m|<1$ for $m,n > N$. This gives $a_n=a_m$ for $m,n > N$, since all $a_n in mathbb Z.$

share|cite|improve this answer

answered Mar 13 at 10:25

FredFred

48.5k11849

$endgroup$

add a comment | 

$begingroup$

If $(a_n)$ is convergent, then $(a_n)$ is Cauchy. Hence there is $N in mathbb N$ such that

$|a_n-a_m|<1$ for $m,n > N$. This gives $a_n=a_m$ for $m,n > N$, since all $a_n in mathbb Z.$

share|cite|improve this answer

answered Mar 13 at 10:25

FredFred

48.5k11849

$endgroup$

share|cite|improve this answer

answered Mar 13 at 10:25

FredFred

48.5k11849

answered Mar 13 at 10:25

FredFred

48.5k11849

answered Mar 13 at 10:25

FredFred

48.5k11849

0

$begingroup$

Option.

Since $a_n in mathbbZ^+$, $n=1,2,3,.....,$ is convergent in $mathbbR$ it is Cauchy.

Let $epsilon <1$ given.

There is a $N$ s.t. for $n ge mge N$

$|a_n-a_m| <1$ , since $a_n,a_m$ are integers we have

$a_n=a_m$.

share|cite|improve this answer

answered Mar 13 at 10:45

Peter SzilasPeter Szilas

11.6k2822

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  is an and am simply referring to the ai and aj in the question?
  $endgroup$
  – john smith
  Mar 13 at 10:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @johnsmith Yes, exactly.
  $endgroup$
  – Pedro Tamaroff♦
  Mar 13 at 10:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  John.Yes! My comment disappeared.Pedro answered already.n,m are dummies.You can rewrite the whole thing with i j, instead.Ok?
  $endgroup$
  – Peter Szilas
  Mar 13 at 10:58
  

  
  
  
  

add a comment | 

0

$begingroup$

Option.

Since $a_n in mathbbZ^+$, $n=1,2,3,.....,$ is convergent in $mathbbR$ it is Cauchy.

Let $epsilon <1$ given.

There is a $N$ s.t. for $n ge mge N$

$|a_n-a_m| <1$ , since $a_n,a_m$ are integers we have

$a_n=a_m$.

share|cite|improve this answer

answered Mar 13 at 10:45

Peter SzilasPeter Szilas

11.6k2822

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  is an and am simply referring to the ai and aj in the question?
  $endgroup$
  – john smith
  Mar 13 at 10:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @johnsmith Yes, exactly.
  $endgroup$
  – Pedro Tamaroff♦
  Mar 13 at 10:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  John.Yes! My comment disappeared.Pedro answered already.n,m are dummies.You can rewrite the whole thing with i j, instead.Ok?
  $endgroup$
  – Peter Szilas
  Mar 13 at 10:58
  

  
  
  
  

add a comment | 

$begingroup$

Option.

Since $a_n in mathbbZ^+$, $n=1,2,3,.....,$ is convergent in $mathbbR$ it is Cauchy.

Let $epsilon <1$ given.

There is a $N$ s.t. for $n ge mge N$

$|a_n-a_m| <1$ , since $a_n,a_m$ are integers we have

$a_n=a_m$.

share|cite|improve this answer

answered Mar 13 at 10:45

Peter SzilasPeter Szilas

11.6k2822

$endgroup$

share|cite|improve this answer

answered Mar 13 at 10:45

Peter SzilasPeter Szilas

11.6k2822

answered Mar 13 at 10:45

Peter SzilasPeter Szilas

11.6k2822

answered Mar 13 at 10:45

Peter SzilasPeter Szilas

11.6k2822