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Math Help - test

  1. #1
    Junior Member
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    Apr 2010
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    test

    <br />
{a}_{n}=\frac{1}{n}\sum_{k=1}^{n}\frac{1}{k}<br />
=\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{1}{\sqrt{n}  }\frac{1}{k}<br />
<\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{1}{{k}^{3/2}}<br />
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  2. #2
    Junior Member
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    Apr 2010
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    Let x,t\in \left(a,b \right).

    By the mean value theorem of derivatives:

    f(x)-f(t)=(x-t)f'(\lambda )

    where \lambda \in (x,t). Hence:

    \left| f(x)-f(t)\right|\leq M\left| x-t\right|

    Now for any \epsilon >0, we can put

    \delta =\frac{\epsilon }{M}

    so that whenever \left| x-t\right|<\delta , we will have \left| f(x)-f(t)\right|<\epsilon . This proves that f(x) is uniformly continuous.
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