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Thread: Sum of divisors proof question

  1. #1
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    Aug 2010
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    Sum of divisors proof question

    s(n) denotes the sum of divisors function

    suppose n=p^k, where p is prime and k is a positive integer
    s(p^k)= [(p^k+1)-1]/(p-1)

    show:
    1+ 1/p < s(n)/n < 1+ 1/(p-1)

    Also if p=3 how large does k have to be to make s(n)/n > 1.4999



    I know that if the above is expanded then it makes sense that each part is less than the next but I'm not sure if this is a proper proof as I am using what they have given and just expanding it and showing it's true. Is there another way to go about it? and I know k has to be something like 15, but is there an easier way to figure it out rather than just starting from k=1 and working my way up?
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  2. #2
    Super Member
    Joined
    Jan 2009
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    Yes , expanding is a proper proof , or we can set $\displaystyle x = 1/p $ and suppose $\displaystyle n $ is a power of a prime .


    Then $\displaystyle \frac{s(n)}{n} = 1+ x + x^2 + ... + x^k $


    The LHS from the inequality is $\displaystyle 1 + x $ while the RHS is $\displaystyle 1 + \frac{x}{1-x} = \frac{1}{1-x} = 1 + x + x^2 + ... $ by combining them we can see what we need to show :


    $\displaystyle 1+x \leq 1 + x + x^2 + ... + x^k < 1 + x + x^2 + .... $ which is obviously true for $\displaystyle k \geq 1 $ .
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