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Math Help - The sum of postive divisors Function

  1. #1
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    The sum of postive divisors Function

    Let n in Z with n>0. Prove that the sum of 1/d = sigma(n)/n.
    d is the divisors n. (d|n,d>o)

    I must be doing something wrong but i need help spotting it.

    For the sum of (1/d) I sub in 1/p1^a1+p2^a2+...Pr^ar.

    Then for sigma(n) I have

    [(p1^(a1+1) - 1)/P1-1] [(P2^(a2+1) - 1)/ P2-1]...[Pr^(ar+1)-1/pr-1]

    and for n i used an arbitrary prime factorization. n=p1^a1P2^a2...pr^ar

    I am not sure if started off right. but my calculations does not prove that statement.

    Did i start the wrong way and if i did not how can I can i compute to prove the statement.
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  2. #2
    o_O
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    Suppose that there are k divisors of n: d_{1}, d_{2}, \ldots, d_{k} such that d_{1}d_{k} = d_{2}d_{k-1} = ... = n

    Then, d_{k} + d_{k-1} + \ldots + d_{2} + d_{1} = \sigma (n)

    Divide both sides by n and you get ...
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