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Prove that the given sequence converges

How do I show that an interleaved sequence converges?Prove if the sequence is bounded & monotonic & convergesConvergence of a sequence $c_n$What about the convergence of the geometric mean sequence of the terms of a given convergent sequence?If a sequence $a_n$ has exactly three partial limits, and a sequence $b_n$ has exactly two partial limits. could the sequence $c_n=a_n+b_n$ converge?Convergence of a sequence and some of its subsequencesHow to find examples of two sequences$c_n$ is a shuffling of $a_n$ and $b_n$. Prove that $c_n$ converges iff $a_n$ and $b_n$ converge to the same number.Prove that the sequence $(a_n)_n=1^infty$ convergesProve that the following sequence converges

0

1

$begingroup$

Prove that the given sequence $a_n$ converges:

$a_1 > 0, a_2 > 0$

$a_n+1 = frac2a_n + a_n-1$ for $n geq 2$

As I observed, this sequence does not seem to be monotonic and that it could be bounded since the values of $a_1$ and $a_2$ are arbitrary positive numbers.

If the limit of the sequence existed, it would be equal to 1 by letting the limit of $a_n$ be x as n goes to infinity, and solving the equation x = $frac2x + x$ => x = 1 or -1, from which we choose x = 1 since x must be positive.

The only idea that came to my mind is bounding the sequence using two other sequences that could be shown to converge to 1 (Let these sequences be $b_n$ and $c_n$):

$b_n <= a_n <= c_n$

If we could find such sequences,and prove that they converge to 1, the problem would be solved. So, I tried to bound the sequence from both sides, and try to show that the limits are equal to 1, but failed to find such sequences. I found that it is a little difficult to analyze sequences of the form presented in the problem since the sequence fluctuates a lot.

I am not sure how to start off, any ideas or tricks for such problems would be appreciated.

convergence

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edited 19 hours ago

Aidyn

asked yesterday

AidynAidyn

161

New contributor

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

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why has this question been downvoted? The OP has clearly shown their efforts.
  $endgroup$
  – Toby Mak
  18 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This was my mistake. This is my first time posting a question here, I did not specify details of my effort, and edited it afterwards.
  $endgroup$
  – Aidyn
  18 hours ago
  

  
  
  
  

add a comment | 

0

1

$begingroup$

Prove that the given sequence $a_n$ converges:

$a_1 > 0, a_2 > 0$

$a_n+1 = frac2a_n + a_n-1$ for $n geq 2$

As I observed, this sequence does not seem to be monotonic and that it could be bounded since the values of $a_1$ and $a_2$ are arbitrary positive numbers.

If the limit of the sequence existed, it would be equal to 1 by letting the limit of $a_n$ be x as n goes to infinity, and solving the equation x = $frac2x + x$ => x = 1 or -1, from which we choose x = 1 since x must be positive.

The only idea that came to my mind is bounding the sequence using two other sequences that could be shown to converge to 1 (Let these sequences be $b_n$ and $c_n$):

$b_n <= a_n <= c_n$

If we could find such sequences,and prove that they converge to 1, the problem would be solved. So, I tried to bound the sequence from both sides, and try to show that the limits are equal to 1, but failed to find such sequences. I found that it is a little difficult to analyze sequences of the form presented in the problem since the sequence fluctuates a lot.

I am not sure how to start off, any ideas or tricks for such problems would be appreciated.

convergence

share|cite|improve this question

edited 19 hours ago

Aidyn

asked yesterday

AidynAidyn

161

New contributor

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

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why has this question been downvoted? The OP has clearly shown their efforts.
  $endgroup$
  – Toby Mak
  18 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This was my mistake. This is my first time posting a question here, I did not specify details of my effort, and edited it afterwards.
  $endgroup$
  – Aidyn
  18 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Prove that the given sequence $a_n$ converges:

$a_1 > 0, a_2 > 0$

$a_n+1 = frac2a_n + a_n-1$ for $n geq 2$

As I observed, this sequence does not seem to be monotonic and that it could be bounded since the values of $a_1$ and $a_2$ are arbitrary positive numbers.

If the limit of the sequence existed, it would be equal to 1 by letting the limit of $a_n$ be x as n goes to infinity, and solving the equation x = $frac2x + x$ => x = 1 or -1, from which we choose x = 1 since x must be positive.

The only idea that came to my mind is bounding the sequence using two other sequences that could be shown to converge to 1 (Let these sequences be $b_n$ and $c_n$):

$b_n <= a_n <= c_n$

If we could find such sequences,and prove that they converge to 1, the problem would be solved. So, I tried to bound the sequence from both sides, and try to show that the limits are equal to 1, but failed to find such sequences. I found that it is a little difficult to analyze sequences of the form presented in the problem since the sequence fluctuates a lot.

I am not sure how to start off, any ideas or tricks for such problems would be appreciated.

convergence

share|cite|improve this question

edited 19 hours ago

Aidyn

asked yesterday

AidynAidyn

161

New contributor

Aidyn 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 19 hours ago

Aidyn

asked yesterday

AidynAidyn

161

New contributor

Aidyn 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 19 hours ago

Aidyn

asked yesterday

AidynAidyn

161

New contributor

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

asked yesterday

AidynAidyn

161

New contributor

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

asked yesterday

AidynAidyn

161