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Thread: Why is this limit true...?

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

    I've just seen this:
    $\displaystyle \{x_i\}_{i=1}^\infty \ \sum _i ^\infty |x_i| < + \infty \rightarrow lim_{+\infty} x_i = 0$
    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:
    $\displaystyle \{x_i\}_{i=1}^\infty \ \sum _i ^\infty |x_i| < + \infty \rightarrow lim_{+\infty} x_i = 0$
    why is this true?

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

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

    $\displaystyle |x_n|$

    $\displaystyle =\ ||x_n||$

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

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

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

    $\displaystyle <\ \frac\epsilon2+\frac\epsilon2$

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

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

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

    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 $\displaystyle \sum _i ^\infty |x_i| < + \infty$ means $\displaystyle S_n = \sum\limits_{k = 1}^n {\left| {x_k } \right|} \, \Rightarrow \,\left( {\exists L} \right)\left[ {\left( {S_n } \right) \to L} \right]$.
    So the sequence $\displaystyle S_n$ is a Cauchy sequence.
    That means we can make $\displaystyle \left| {x_n } \right| = \left| {S_n - S_{n - 1} } \right| < \varepsilon $.
    That is sufficient to prove the property: $\displaystyle \left| {x_n } \right| \to 0$
    So proscientia's post shows the sequence is Cauchy, right?
    And as you've stated the distance $\displaystyle |S_n - S_{n-1}| \ is <\epsilon $ but since $\displaystyle \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 $\displaystyle |S_n - S_{n-1}| \ is <\epsilon $ but since $\displaystyle \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 $\displaystyle S_n$ equals the limit of $\displaystyle S_{n-1}$ and since, as I wrote in the first message, $\displaystyle |x_n|=S_n-S_{n-1}$, then...

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

    Tonio
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