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Math Help - Another Series: Looking at conv vs. div

  1. #1
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    Another Series: Looking at conv vs. div

    So we have summation of 1/[k(k+1)(k+2)]

    We know that 1/[k(k+1)(k+2)] < 1/[(k+1)(k+2)] which I know converges.

    Thus, 1/[k(k+1)(k+2)] also converges.

    Can someone check my reasoning?
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  2. #2
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    Quote Originally Posted by WartonMorton View Post
    So we have summation of 1/[k(k+1)(k+2)]

    We know that 1/[k(k+1)(k+2)] < 1/[(k+1)(k+2)] which I know converges.

    Thus, 1/[k(k+1)(k+2)] also converges.

    Can someone check my reasoning?

    I think this is fine. (Just make sure you have the proof that 1/[(k+1)(k+2)] converges).
    As an exercise you can try to formally prove it using your argument above and the definition of convergence. It will give you good experience.


    (Also you may want to put these/related questions in the analysis sub-forum.)
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