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Math Help - Bounds on sum of reciprocal of squares

  1. #1
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    Bounds on sum of reciprocal of squares

    I would like to have a bound on the on the following sum which for me will do but am not able to prove it. Can someone help please?

    Is it true: \forall m>1, we have

    \sum_{n>m}\frac{1}{n^2}<\frac{1}{m-1}?
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  2. #2
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    Quote Originally Posted by Abegunde View Post
    I would like to have a bound on the on the following sum which for me will do but am not able to prove it. Can someone help please?

    Is it true: \forall m>1, we have

    \sum_{n>m}\frac{1}{n^2}<\frac{1}{m-1}?

    \displaystyle{\sum\limits^\infty_{n>m}\frac{1}{n^2  }\leq \sum\limits^\infty_{n>m}\frac{1}{n(n-1)}=\sum^\infty_{n>m}\left(\frac{1}{n-1}-\frac{1}{n}\right) ....

    Tonio
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