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Math Help - [SOLVED] Sum of divisors

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    MHF Contributor Bruno J.'s Avatar
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    [SOLVED] Sum of divisors

    Let k denote the number of positive divisors of the positive integer n, and S the sum of the positive divisors of n. Show that

    k\sqrt{n} < S < \sqrt{2k}n
    Last edited by Bruno J.; May 18th 2010 at 02:17 PM.
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    Quote Originally Posted by Bruno J. View Post
    Let k denote the number of positive divisors of the positive integer n, and S the sum of the positive divisors of n. Show that

    k\sqrt{n} < S < \sqrt{2k}n
    we also need to have n > 1. recall that for any distinct (real) numbers x_j > 0 we have (x_1 + \cdots + x_k)\left( \frac{1}{x_1} + \cdots + \frac{1}{x_k} \right) > k^2. thus:

    S^2=\sum_{d \mid n}d \cdot \sum_{d \mid n} \frac{n}{d}=n \sum_{d \mid n} d \cdot \sum_{d \mid n} \frac{1}{d} > nk^2.

    to prove the other inequality first note that since e^x > 1 + \frac{x^2}{2} we have x > \ln \left(1 + \frac{x^2}{2} \right) and so \sqrt{2k} > \ln(k+1). using this we have:

    \frac{1}{n}S=\sum_{d \mid n} \frac{1}{d} \leq \sum_{j=1}^k \frac{1}{j} < \int_1^{k+1} \frac{dx}{x}=\ln(k+1) < \sqrt{2k}.
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    MHF Contributor Bruno J.'s Avatar
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    Great job! (Thanks for specifying n>1 ).

    Here's how I proved the first inequality; we have S = \frac{1}{2}\sum_{d|n}\left(d+\frac{n}{d}\right), and \frac{1}{2}\left(d+\frac{n}{d}\right) \geq \sqrt n by the arithmetic/geometric mean inequality.
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    Quote Originally Posted by Bruno J. View Post

    Here's how I proved the first inequality; we have S = \frac{1}{2}\sum_{d|n}\left(d+\frac{n}{d}\right), and \frac{1}{2}\left(d+\frac{n}{d}\right) \geq \sqrt n by the arithmetic/geometric mean inequality.
    oh yeah, this is easier!
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