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Math Help - Sequence of partial sums is Cauchy

  1. #1
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    Sequence of partial sums is Cauchy

    If (x_j) is a sequence in an inner product space X, such that the series
    \|x_1\| + \|x_2\| + \cdots converges.

    Show that (s_n) is a Cauchy sequence , where s_n = x_1 + \cdots + x_n.
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by frater_cp View Post
    If (x_j) is a sequence in an inner product space X, such that the series
    \|x_1\| + \|x_2\| + \cdots converges.

    Show that (s_n) is a Cauchy sequence , where s_n = x_1 + \cdots + x_n.
    Let S be the limit of T_n=\sum_{j=1}^n \|x_j\|

    Hence, \forall \epsilon>0,\exists N\in\mathbb{N},\forall k\geq N, \|T_k-S\|<\frac\epsilon 2
    In particular, we have :
    \begin{array}{ll} \forall m\geq N ~:~ \|T_m-S\|<\frac\epsilon 2 \\ \forall n\geq N ~:~ \|T_n-S\|<\frac\epsilon 2\end{array}

    So \|T_n-S\|+\|T_m-S\|<\epsilon
    Which is equivalent to \|T_n-S\|+\|S-T_m\|<\epsilon

    Since \|.\| is a norm, it satisfies the triangle inequality :

    \|T_n-T_m\|=\|T_n-S+S-T_m\|\leq \|T_n-S\|+\|S-T_m\|<\epsilon

    But (assuming n>m) \|T_n-T_m\|=\left\|\|x_n\|+\|x_{n-1}\|+\dots+\|x_{m+1}\|\right\|=a \left(\|x_n\|+\|x_{n-1}\|+\dots+\|x_{m+1}\|\right), where a is the norm of the unit vector.

    Again, by the triangle inequality, we know that \|s_n-s_m\|=\|x_n+x_{n-1}+\dots+x_{m+1}\|\leq \|x_n\|+\|x_{n-1}\|+\dots+\|x_{m+1}\|

    So finally, we have (letting \epsilon'=\epsilon/a)

    \forall \epsilon'>0,\exists N\in\mathbb{N},\forall n,m\geq N, \|s_n-s_m\|<\epsilon'

    And this proves that (s_n) is a Cauchy sequence
    Last edited by Moo; May 31st 2009 at 01:19 PM. Reason: e/a instead of ae
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  3. #3
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    want to check something

    Hi Moo.

    Thank you very much for your elegant answer.

    I want to check something in your solution I do not understand.

    Its were you state an equality (where assumed n>m) and then you have:
    a = the norm of the unit vector

    I do not understand this equality can you explain where this whole line comes from please??

    Kind Regards!
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  4. #4
    Moo
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    Hi
    Quote Originally Posted by frater_cp View Post
    Hi Moo.

    Thank you very much for your elegant answer.

    I want to check something in your solution I do not understand.

    Its were you state an equality (where assumed n>m) and then you have:
    a = the norm of the unit vector

    I do not understand this equality can you explain where this whole line comes from please??

    Kind Regards!
    No problem. This proves that you're following what I've been doing...

    \left\|\|x_n\|+\|x_{n-1}\|+\dots+\|x_{m+1}\|\right\|=a \left(\|x_n\|+\|x_{n-1}\|+\dots+\|x_{m+1}\|\right)
    ...wrong

    I'm sorry, I guess I was confusing myself... The norm of a number is its absolute value.
    Here, \|x_n\|+\|x_{n-1}\|+\dots+\|x_{m+1}\| is a number, a positive number, since a norm is always positive.

    So \left\|\|x_n\|+\|x_{n-1}\|+\dots+\|x_{m+1}\|\right\|=\|x_n\|+\|x_{n-1}\|+\dots+\|x_{m+1}\|

    And in the final steps, you juste have to keep \epsilon instead of \epsilon'

    Again, excuse me for my mistake
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