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Thread: test

  1. #1
    Junior Member
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    Apr 2010
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    test

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

    By the mean value theorem of derivatives:

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

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

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

    Now for any $\displaystyle \epsilon >0$, we can put

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

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