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Math Help - Sequences

  1. #1
    Member roshanhero's Avatar
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    Sequences

    If a real sequence converges,then it is bounded.
    How can we prove this statement?
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  2. #2
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    Let its limit be L. The very definition of convergence implies the existence of a natural number N such that |a_{n}-L|<1 for every n \geq N. The former inequality implies in turn that |a_{n}| < 1+|L| for every n \geq N.

    Hence, starting from the N-th element of the sequence you can place them all inside of a disk of radius 1+|L| (and centered at the origin). it's clear that up to this moment some elements of the sequence might remain outside of this disk, but since it's only a finite number of them, we are done, right?
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  3. #3
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    I think if we choose :

    K = max{  |a_{1}|,|a_{2}|.........|a_{N}|,|L| +1} then we will have for all nεN  |a_{n}|\leq K
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  4. #4
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    Quote Originally Posted by xalk View Post
    I think if we choose :

    K = max{  |a_{1}|,|a_{2}|.........|a_{N}|,|L| +1} then we will have for all nεN  |a_{n}|\leq K
    A little clarification: N is the positive integer such that if n> N, then |a_n- L|< 1 which is guarenteed to exist since L is the limit of the sequence.
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    A little clarification: N is the positive integer such that if n> N, then |a_n- L|< 1 which is guarenteed to exist since L is the limit of the sequence.
    Well ,i think coquitao clarified that in his post,did he not?
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  6. #6
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    Quote Originally Posted by coquitao View Post
    Let its limit be L. The very definition of convergence implies the existence of a natural number N such that |a_{n}-L|<1 for every n \geq N. The former inequality implies in turn that |a_{n}| < 1+|L| for every n \geq N.

    Hence, starting from the N-th element of the sequence you can place them all inside of a disk of radius 1+|L| (and centered at the origin). it's clear that up to this moment some elements of the sequence might remain outside of this disk, but since it's only a finite number of them, we are done, right?
    How do you get from |a_{n}-L|<1 to |a_{n}| < 1+|L|? You might need some explanation to that. Could use reverse triangle inequality to show that |a_n-L|\geq |a_n|-|L|
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  7. #7
    Super Member redsoxfan325's Avatar
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    |a_n|=|a_n-L+L|\leq|a_n-L|+|L|\implies|a_n|-|L|\leq|a_n-L|

    Using this you have |a_n|-|L|\leq|a_n-L|<1\implies |a_n|-|L|<1\implies |a_n|<1+|L|
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