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Math Help - Why is this limit true...?

  1. #1
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    Why is this limit true...?

    I've just seen this:
    \{x_i\}_{i=1}^\infty \ \sum _i ^\infty |x_i| < + \infty \rightarrow lim_{+\infty} x_i = 0
    why is this true?
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  2. #2
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    Quote Originally Posted by bigdoggy View Post
    I've just seen this:
    \{x_i\}_{i=1}^\infty \ \sum _i ^\infty |x_i| < + \infty \rightarrow lim_{+\infty} x_i = 0
    why is this true?

    Put\,\,S:=\sum\limits_{i=1}^\infty|x_i| \Longrightarrow since\,\,|x_n|=\sum\limits_{i=1}^n |x_i|-\sum\limits_{i=1}^{n-1}|x_i| , passing to the limit when n\rightarrow \infty gives the desired result.

    Tonio
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  3. #3
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    Let \sum_{i\,=\,1}^\infty|x_i|=L. Then given \epsilon>0, \exists\,N\in\mathbb N such that n\geqslant N\ \Rightarrow\ \left|\sum_{i\,=\,1}^n|x_i|-L\right|<\frac\epsilon2. Hence, \forall\,n\geqslant N+1,

    |x_n|

    =\ ||x_n||

    =\ \left|\sum_{i\,=\,1}^n|x_i|-\sum_{i\,=\,1}^{n-1}|x_i|\right|

    =\ \left|\left(\sum_{i\,=\,1}^n|x_i|-L\right)-\left(\sum_{i\,=\,1}^{n-1}|x_i|-L\right)\right|

    \leqslant\ \left|\sum_{i\,=\,1}^n|x_i|-L\right|+\left|\sum_{i\,=\,1}^{n-1}|x_i|-L\right|

    <\ \frac\epsilon2+\frac\epsilon2

    =\ \epsilon
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  4. #4
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    Quote Originally Posted by tonio View Post
    Put\,\,S:=\sum\limits_{i=1}^\infty|x_i| \Longrightarrow since\,\,|x_n|=\sum\limits_{i=1}^n |x_i|-\sum\limits_{i=1}^{n-1}|x_i| , passing to the limit when n\rightarrow \infty gives the desired result.

    Tonio
    Please could you expand on this....why have you separated the |x_n| and how do I proceed after?
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  5. #5
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    Quote Originally Posted by bigdoggy View Post
    Please could you expand on this....why have you separated the |x_n| and how do I proceed after?
    Quote Originally Posted by bigdoggy View Post
    I've just seen this:
    \{x_i\}_{i=1}^\infty \ \sum _i ^\infty |x_i| < + \infty \rightarrow lim_{+\infty} x_i = 0 why is this true?
    To say that  \sum _i ^\infty |x_i| < + \infty means S_n  = \sum\limits_{k = 1}^n {\left| {x_k } \right|} \, \Rightarrow \,\left( {\exists L} \right)\left[ {\left( {S_n } \right) \to L} \right].
    So the sequence S_n is a Cauchy sequence.
    That means we can make \left| {x_n } \right| = \left| {S_n  - S_{n - 1} } \right| < \varepsilon .
    That is sufficient to prove the property: \left| {x_n } \right| \to 0
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  6. #6
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    Quote Originally Posted by bigdoggy View Post
    Please could you expand on this....why have you separated the |x_n| and how do I proceed after?
    Do exactly as I wrote: take the limit in both sides when n\rightarrow \infty...
    I represented (not separated) x_n as I did above so that we'll be able to evaluate the sequence's limit.

    Tonio
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  7. #7
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    Quote Originally Posted by tonio View Post
    Do exactly as I wrote: take the limit in both sides when n\rightarrow \infty...
    I represented (not separated) x_n as I did above so that we'll be able to evaluate the sequence's limit.

    Tonio
    So, evaluate lim_{n \rightarrow \infty} \sum ^n _1x_i -  lim_{n \rightarrow \infty} \sum^{n-1} _1 x_i

    Am I ok to drop the modulus' when evaluating the limit as above?
    I'm not sure how to evaluate this
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  8. #8
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    Quote Originally Posted by Plato View Post
    To say that  \sum _i ^\infty |x_i| < + \infty means S_n  = \sum\limits_{k = 1}^n {\left| {x_k } \right|} \, \Rightarrow \,\left( {\exists L} \right)\left[ {\left( {S_n } \right) \to L} \right].
    So the sequence S_n is a Cauchy sequence.
    That means we can make \left| {x_n } \right| = \left| {S_n  - S_{n - 1} } \right| < \varepsilon .
    That is sufficient to prove the property: \left| {x_n } \right| \to 0
    So proscientia's post shows the sequence is Cauchy, right?
    And as you've stated the distance |S_n - S_{n-1}| \ is <\epsilon but since \epsilon >0 as in a Cauchy sequence, that implies it must tend to zero....is my thinking right?
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  9. #9
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    Quote Originally Posted by bigdoggy View Post
    So proscientia's post shows the sequence is Cauchy, right?
    And as you've stated the distance |S_n - S_{n-1}| \ is <\epsilon but since \epsilon >0 as in a Cauchy sequence, that implies it must tend to zero....is my thinking right?

    No, but close: since the sequence (of partial sums of the series) is Cauchy then it has a limit, but since the limit of  S_n equals the limit of S_{n-1} and since, as I wrote in the first message, |x_n|=S_n-S_{n-1}, then...

    Be careful: S=\lim_{n\rightarrow \infty}S_n can NOT be zero unless x_i=0\,\,\forall\,i !

    Tonio
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