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Math Help - Proving an inequality (Area under graph)

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    Proving an inequality (Area under graph)

    Given that y=\frac{1}{x^2}, where x>0.
    By considering the area under the curve and the areas of rectangles, for n \le x \le n+1, where n \ge 1 is an integer, show that

    \frac{n^2+n+1}{(n+1)^2}<1+\frac{1}{2^2}+\frac{1}{3  ^2}+...+\frac{1}{(n+1)^2}<\frac{2n+1}{n+1}



    I started with \frac{1}{(n+1)^2}<\int^{n+1}_{n} \frac{1}{x^2}dx<\frac{1}{n^2} which leads to
    \frac{1}{(n+1)^2}<\frac{1}{n(n+1)}<\frac{1}{n^2}

    Anyone can help? thanks!
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  2. #2
    MHF Contributor red_dog's Avatar
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    We have \frac{1}{(k+1)^2}<\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}

    1=1

    \frac{1}{2^2}<1-\frac{1}{2}

    \frac{1}{3^2}<\frac{1}{2}-\frac{1}{3}

    .................................................. ......................

    \frac{1}{n^2}<\frac{1}{n}-\frac{1}{n+1}

    Summing the inequalities we have

    1+\frac{1}{2^2}+\ldots+\frac{1}{(n+1)^2}<2-\frac{1}{n+1}=\frac{2n+1}{n+1}

    Now, \frac{n^2+n+1}{(n+1)^2}=\frac{n}{n+1}+\frac{1}{(n+  1)^2} and the left side inequality can be written as

    \frac{n}{n+1}<1+\frac{1}{2^2}+\frac{1}{3^2}+\ldots  +\frac{1}{n^2}

    We have:

    \frac{1}{1\cdot2}<1

    \frac{1}{2\cdot3}<\frac{1}{2^2}

    .................................................. ........

    \frac{1}{n(n+1)}<\frac{1}{n^2}

    Adding the inequalties we get the result.
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