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    Flicking through an old book I came across this problem

    By comparing the area under the graph of 1/(x^2) between x - r-1 and r with thw area of an appropriate rectangle over the same interval

    show that (1/(r^2) < integral between (r-1) and r of 1/(x^2) DONE THAT
    Deduce that sum between r=1 and N of 1/(r^2) < integral between (1) and N of 1/(x^2) DONE THAT

    But can't do the following

    Hence show that sum between r=1 and N of 1/(r^2) < 2

    Help please
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    (Original post by maggiehodgson)
    Flicking through an old book I came across this problem

    By comparing the area under the graph of 1/(x^2) between x - r-1 and r with thw area of an appropriate rectangle over the same interval

    show that (1/(r^2) < integral between (r-1) and r of 1/(x^2) DONE THAT
    Deduce that sum between r=1 and N of 1/(r^2) < integral between (1) and N of 1/(x^2) DONE THAT

    But can't do the following

    Hence show that sum between r=1 and N of 1/(r^2) < 2

    Help please
    Show that the integral is always less than 2 regardless of what the value of N is.
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    That's the bit I can't do.

    I get, for the integral, 1/(r(r-1)) and then what?
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    (Original post by maggiehodgson)
    That's the bit I can't do.

    I get, for the integral, 1/(r(r-1)) and then what?
    You should get the integral as -1/x and then sub in the limits of N and 1 and then it should be clear.
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    if the integral is 2 and the rectangle sum is less than the integral then the rectangle sum is less than 2
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    BUT I've got to show that

    the integral is < 2 regardless of the value of N. don't think I can say "let the integral = 2" as a starting point.
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    A sketch of the curve and thinking about the summation as rectangles and the integral as the total area under the curve gives the following inequality  \displaystyle \sum_{r=2}^n 1/r^2 \leq \int_1^n1/x^2 \ dx =1-1/n&lt;1 .
    Now this inequality involves  \sum_{r=2}^n1/r^2 but you want the inequality for  \sum_{r=1}^n 1/r^2 and it's just a one step manipulation which gives the result they want.
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    (Original post by B_9710)
    A sketch of the curve and thinking about the summation as rectangles and the integral as the total area under the curve gives the following inequality  \displaystyle \sum_{r=2}^n 1/r^2 \leq \int_1^n1/x^2 \ dx =1-1/n&lt;1 .
    Now this inequality involves  \sum_{r=2}^n1/r^2 but you want the inequality for  \sum_{r=1}^n 1/r^2 and it's just a one step manipulation which gives the result they want.
    Thanks

    I think that has sorted me out.
 
 
 
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