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Math Help - Inequalities

  1. #1
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    Inequalities

    Prove that  \frac{\sigma{(n)}}{n} < \frac{n}{\phi(n)} < \frac{\pi^{2}}{6} \frac{\sigma(n)}{n} if n \geq 2
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  2. #2
    Super Member PaulRS's Avatar
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    For each prime p let \alpha_p(n) be the greatest k\in \mathbb{N} such that p^k|n.

    Then \sigma(n)=\prod_{p|n}\left(1+p+...+p^{\alpha_p(n)}  \right) = n\cdot \prod_{p|n}\left(1+p^{-1}+...+p^{-\alpha_p(n)}\right) =n\cdot \prod_{p|n}\left(\frac{1-p^{-(1+\alpha_p(n))}}{1-p^{-1}}\right)

    But remember that \phi(n)=n\cdot \prod_{p|n}\left(1-p^{-1}\right) thus: \sigma(n)\cdot \phi(n) = n^2\cdot \prod_{p|n}\left(1-p^{-(1+\alpha_p(n))}\right) .

    Now: \tfrac{6}{\pi^2} = \prod_{p \text{ prime}}\left(1-p^{-2}\right) < \prod_{p|n}\left(1-p^{-(1+\alpha_p(n))}\right) = \tfrac{\sigma(n)\cdot \phi(n)}{n^2} \leq 1 - equality holds on the RHS iff n = 1 -

    The right inequality is simple enough, each factor is less than 1 - and there are no factors only if n = 1-, now for the left inequality 1-p^{-2} \leq 1-p^{-(1+\alpha_p(n))} for each p|n, and 1-p^{-2} < 1 when p\not | n -since n can't have all the primes as divisors, it follows that some inequality must be strict.
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