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Math Help - Infinite series (Analysis)

  1. #1
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    Infinite series (Analysis)

    Hey guys,

    How do I prove that (1/nx) --> 0 pointwise but not uniformly on (0,1) as n tends to infinity.

    thanks in advance, I appreciate your help.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by thahachaina View Post
    Hey guys,

    How do I prove that (1/nx) --> 0 pointwise but not uniformly on (0,1) as n tends to infinity.

    thanks in advance, I appreciate your help.
    On (0,1) and any \varepsilon>0

     <br />
\frac{1}{nx}<\varepsilon,\ \ \forall n>N=\left \lceil \frac{1}{\varepsilon x}\right \rceil<br />

    which proves pointwise convergence.

    For uniform convergence for \varepsilon>0 we need to find an N such that for all x\in (0,1):

     <br />
\frac{1}{nx}<\varepsilon,\ \ \forall n>N

    But for x<\frac{1}{n\varepsilon} where n>N;\ \frac{1}{nx}>\varepsilon which contradicts what would have to be true for uniform convergence to hold.

    CB
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  3. #3
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    Re: Infinite series (Analysis)

    I'm lost again! What is < br >?
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  4. #4
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    Re: Infinite series (Analysis)

    I think that's just a formatting code that's no longer used - nothing mathematical about it. Here's CB's reply without the "br" formatting codes:

    -------------------------

    On (0,1) and any \varepsilon>0,

    \frac{1}{nx}<\varepsilon,\ \ \forall n>N=\left \lceil \frac{1}{\varepsilon x}\right \rceil

    which proves pointwise convergence.

    For uniform convergence for \varepsilon>0 we need to find an N such that for all x\in (0,1):

    \frac{1}{nx}<\varepsilon,\ \ \forall n>N

    But for x<\frac{1}{n\varepsilon} where n>N;\ \frac{1}{nx}>\varepsilon which contradicts what would have to be true for uniform convergence to hold.

    CB

    -------------------------

    By the way, why did you resurrect this 4-year-old thread?

    - Hollywood
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