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Math Help - convergent or divergent series?

  1. #1
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    convergent or divergent series?

    [sqrt(k) - sqrt(k-1)]^k
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  2. #2
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    Quote Originally Posted by WartonMorton View Post
    [sqrt(k) - sqrt(k-1)]^k



    \sqrt{k}-\sqrt{k-1}=\frac{1}{\sqrt{k}+\sqrt{k+1}} , and now raise all to the k-th power...

    Tonio
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  3. #3
    Super Member Failure's Avatar
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    Quote Originally Posted by WartonMorton View Post
    [sqrt(k) - sqrt(k-1)]^k
    Let's see

    0 \leq \left[\sqrt{k}-\sqrt{k-1}\right]^k=\left[\frac{1}{\sqrt{k}+\sqrt{k-1}}\right]^k\leq \left[\frac{1}{2\sqrt{k-1}}\right]^k\leq \frac{1}{2^k}, for k\geq 2

    so, yes, that series converges.
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