No convergent sequence containing infinitely many ones can have a limit other than one.

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Is the following argument correct?

Proposition. There cannot exist a sequence with an infinite number of ones that converges to a limit not equal to one.

Proof. Let $(a_n)$ be a sequence containing an infinite number of ones such that $(a_n)to L$ where $Lneq 1$, then $|1-L|>0$ and from hypothesis there exists an $NinmathbfN$ such that $|a_n-L|<|1-L|/2,forall nge N$.

In addition there must exist a $kinN,N+1,N+2,dots$ such that $a_k=1$, otherwise the sequence may only have finitely many ones, but then $|a_k-L| = |1-L|<|1-L|/2$ a contradiction.

$blacksquare$

real-analysis sequences-and-series proof-verification

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edited Jul 19 at 9:45

asked Jul 19 at 8:21

Atif Farooq

2,7092824

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  Your proof is correct. It is simpler to note that if a sequence converges to a limit $L$ then any subsequence also converges to the same limit.
  – Kavi Rama Murthy
  Jul 19 at 8:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KaviRamaMurthy Thanks.In my defence the text i am using has not introduced subsequences at this point but your remark is greatly appreciated.
  – Atif Farooq
  Jul 19 at 8:26
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

Is the following argument correct?

Proposition. There cannot exist a sequence with an infinite number of ones that converges to a limit not equal to one.

Proof. Let $(a_n)$ be a sequence containing an infinite number of ones such that $(a_n)to L$ where $Lneq 1$, then $|1-L|>0$ and from hypothesis there exists an $NinmathbfN$ such that $|a_n-L|<|1-L|/2,forall nge N$.

In addition there must exist a $kinN,N+1,N+2,dots$ such that $a_k=1$, otherwise the sequence may only have finitely many ones, but then $|a_k-L| = |1-L|<|1-L|/2$ a contradiction.

$blacksquare$

real-analysis sequences-and-series proof-verification

share|cite|improve this question

edited Jul 19 at 9:45

asked Jul 19 at 8:21

Atif Farooq

2,7092824

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Your proof is correct. It is simpler to note that if a sequence converges to a limit $L$ then any subsequence also converges to the same limit.
  – Kavi Rama Murthy
  Jul 19 at 8:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KaviRamaMurthy Thanks.In my defence the text i am using has not introduced subsequences at this point but your remark is greatly appreciated.
  – Atif Farooq
  Jul 19 at 8:26
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Is the following argument correct?

Proposition. There cannot exist a sequence with an infinite number of ones that converges to a limit not equal to one.

Proof. Let $(a_n)$ be a sequence containing an infinite number of ones such that $(a_n)to L$ where $Lneq 1$, then $|1-L|>0$ and from hypothesis there exists an $NinmathbfN$ such that $|a_n-L|<|1-L|/2,forall nge N$.

In addition there must exist a $kinN,N+1,N+2,dots$ such that $a_k=1$, otherwise the sequence may only have finitely many ones, but then $|a_k-L| = |1-L|<|1-L|/2$ a contradiction.

$blacksquare$

real-analysis sequences-and-series proof-verification

share|cite|improve this question

edited Jul 19 at 9:45

asked Jul 19 at 8:21

Atif Farooq

2,7092824

Is the following argument correct?

Proposition. There cannot exist a sequence with an infinite number of ones that converges to a limit not equal to one.

Proof. Let $(a_n)$ be a sequence containing an infinite number of ones such that $(a_n)to L$ where $Lneq 1$, then $|1-L|>0$ and from hypothesis there exists an $NinmathbfN$ such that $|a_n-L|<|1-L|/2,forall nge N$.

In addition there must exist a $kinN,N+1,N+2,dots$ such that $a_k=1$, otherwise the sequence may only have finitely many ones, but then $|a_k-L| = |1-L|<|1-L|/2$ a contradiction.

$blacksquare$

real-analysis sequences-and-series proof-verification

share|cite|improve this question

edited Jul 19 at 9:45

asked Jul 19 at 8:21

Atif Farooq

2,7092824

share|cite|improve this question

share|cite|improve this question

edited Jul 19 at 9:45

edited Jul 19 at 9:45

edited Jul 19 at 9:45

asked Jul 19 at 8:21

Atif Farooq

2,7092824

asked Jul 19 at 8:21

Atif Farooq

2,7092824

asked Jul 19 at 8:21

Atif Farooq

2,7092824

2,7092824

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Your proof is correct. It is simpler to note that if a sequence converges to a limit $L$ then any subsequence also converges to the same limit.
  – Kavi Rama Murthy
  Jul 19 at 8:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KaviRamaMurthy Thanks.In my defence the text i am using has not introduced subsequences at this point but your remark is greatly appreciated.
  – Atif Farooq
  Jul 19 at 8:26
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Your proof is correct. It is simpler to note that if a sequence converges to a limit $L$ then any subsequence also converges to the same limit.
  – Kavi Rama Murthy
  Jul 19 at 8:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KaviRamaMurthy Thanks.In my defence the text i am using has not introduced subsequences at this point but your remark is greatly appreciated.
  – Atif Farooq
  Jul 19 at 8:26
  

  
  
  
  

2

2

Your proof is correct. It is simpler to note that if a sequence converges to a limit $L$ then any subsequence also converges to the same limit.
– Kavi Rama Murthy
Jul 19 at 8:25

Your proof is correct. It is simpler to note that if a sequence converges to a limit $L$ then any subsequence also converges to the same limit.
– Kavi Rama Murthy
Jul 19 at 8:25

@KaviRamaMurthy Thanks.In my defence the text i am using has not introduced subsequences at this point but your remark is greatly appreciated.
– Atif Farooq
Jul 19 at 8:26

@KaviRamaMurthy Thanks.In my defence the text i am using has not introduced subsequences at this point but your remark is greatly appreciated.
– Atif Farooq
Jul 19 at 8:26

add a comment | 

1 Answer
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accepted

The proof is correct. You can generalize it as follows:

Suppose the sequence $(a_n)$ is convergent to $L$. If $bne L$, then the set $ninmathbbN:a_n=b$ is finite.

Proof. Suppose $b<L$ and set $varepsilon=(L-b)/2$. Then there exists $N$ such that $|a_n-L|<varepsilon$, for every $nge N$. This implies
$$
-fracL-b2+L<a_n<fracL-b2+L
$$
in particular $a_n>(L+b)/2>b$, for every $nge N$.

Similarly for $b>L$. QED

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answered Jul 19 at 10:06

egreg

164k1180187

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

The proof is correct. You can generalize it as follows:

Suppose the sequence $(a_n)$ is convergent to $L$. If $bne L$, then the set $ninmathbbN:a_n=b$ is finite.

Proof. Suppose $b<L$ and set $varepsilon=(L-b)/2$. Then there exists $N$ such that $|a_n-L|<varepsilon$, for every $nge N$. This implies
$$
-fracL-b2+L<a_n<fracL-b2+L
$$
in particular $a_n>(L+b)/2>b$, for every $nge N$.

Similarly for $b>L$. QED

share|cite|improve this answer

answered Jul 19 at 10:06

egreg

164k1180187

add a comment | 

up vote
2
down vote



accepted

The proof is correct. You can generalize it as follows:

Suppose the sequence $(a_n)$ is convergent to $L$. If $bne L$, then the set $ninmathbbN:a_n=b$ is finite.

Proof. Suppose $b<L$ and set $varepsilon=(L-b)/2$. Then there exists $N$ such that $|a_n-L|<varepsilon$, for every $nge N$. This implies
$$
-fracL-b2+L<a_n<fracL-b2+L
$$
in particular $a_n>(L+b)/2>b$, for every $nge N$.

Similarly for $b>L$. QED

share|cite|improve this answer

answered Jul 19 at 10:06

egreg

164k1180187

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

The proof is correct. You can generalize it as follows:

Suppose the sequence $(a_n)$ is convergent to $L$. If $bne L$, then the set $ninmathbbN:a_n=b$ is finite.

Proof. Suppose $b<L$ and set $varepsilon=(L-b)/2$. Then there exists $N$ such that $|a_n-L|<varepsilon$, for every $nge N$. This implies
$$
-fracL-b2+L<a_n<fracL-b2+L
$$
in particular $a_n>(L+b)/2>b$, for every $nge N$.

Similarly for $b>L$. QED

share|cite|improve this answer

answered Jul 19 at 10:06

egreg

164k1180187

The proof is correct. You can generalize it as follows:

Suppose the sequence $(a_n)$ is convergent to $L$. If $bne L$, then the set $ninmathbbN:a_n=b$ is finite.

Proof. Suppose $b<L$ and set $varepsilon=(L-b)/2$. Then there exists $N$ such that $|a_n-L|<varepsilon$, for every $nge N$. This implies
$$
-fracL-b2+L<a_n<fracL-b2+L
$$
in particular $a_n>(L+b)/2>b$, for every $nge N$.

Similarly for $b>L$. QED

share|cite|improve this answer

answered Jul 19 at 10:06

egreg

164k1180187

share|cite|improve this answer

share|cite|improve this answer

answered Jul 19 at 10:06

egreg

164k1180187

answered Jul 19 at 10:06

egreg

164k1180187

answered Jul 19 at 10:06

egreg

164k1180187

164k1180187

add a comment | 

add a comment | 

 

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