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Thread: number theory proof

  1. #1
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    number theory proof

    Let with n>1. Prove that 1 + 1/2 + 1/3 + .....1/n
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  2. #2
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    number theory proof

    Let with n>1. Prove that 1 + 1/2 + 1/3 + .....1/n
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  3. #3
    Super Member PaulRS's Avatar
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    Quote Originally Posted by rmpatel5 View Post
    Let with n>1. Prove that 1 + 1/2 + 1/3 + .....1/n
    Let $\displaystyle
    p_n
    $ be the greatest prime not exceeding $\displaystyle n$.

    So we must have: $\displaystyle
    p_n \leqslant n < 2 \cdot p_n
    $ (1) . Suppose this were false, then $\displaystyle
    p_n \leqslant 2 \cdot p_n \leqslant n
    $ but, by Bertrand's Postulate there's a prime $\displaystyle p$ such that $\displaystyle
    p_n < p < 2 \cdot p_n
    $ which is absurd since $\displaystyle
    p_n
    $ is the greatest prime not exceeding $\displaystyle n$.

    Now suppose $\displaystyle
    {1 + \tfrac{1}
    {2} + ... + \tfrac{1}
    {{p_n }} + ... + \tfrac{1}
    {n}} \in
    \mathbb{Z}^ +

    $ then $\displaystyle a=
    \tfrac{{n!}}
    {{p_n }} \cdot \left( {1 + \tfrac{1}
    {2} + ... + \tfrac{1}
    {{p_n }} + ... + \tfrac{1}
    {n}} \right) \in \mathbb{Z}^ +
    $ since $\displaystyle
    \tfrac{{n!}}
    {{p_n }}
    $ is an integer

    $\displaystyle
    a = \underbrace {\tfrac{{n!}}
    {{p_n }} \cdot \tfrac{1}
    {1} + \tfrac{{n!}}
    {{p_n }} \cdot \tfrac{1}
    {2} + ...}_{ \in \mathbb{Z}} + \tfrac{{n!}}
    {{p_n }} \cdot \tfrac{1}
    {{p_n }} + \underbrace {... + \tfrac{{n!}}
    {{p_n }} \cdot \tfrac{1}
    {n}}_{ \in \mathbb{Z}}
    $

    By (1) we have that $\displaystyle
    p_n ^2
    $ doesn't divide $\displaystyle n!$ so $\displaystyle a$ is not an integer. Absurd
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  4. #4
    Super Member Aryth's Avatar
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    There is a classical proof provided by Nicole Oresme, where you consider the subsequence $\displaystyle \{H_{2^k}\}^{\infty}_{k=0}$:

    $\displaystyle H_1 = 1 = 1 + 0\left(\frac{1}{2}\right)$

    $\displaystyle H_2 = 1 + \frac{1}{2} = 1 + 1\left(\frac{1}{2}\right)$

    $\displaystyle H_4 = 1 + \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{4}\right)$

    $\displaystyle > 1 + \frac{1}{2} + \left(\frac{1}{4} + \frac{1}{4}\right) = 1 + 2\left(\frac{1}{2}\right)$

    $\displaystyle H_8 = 1 + \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{4}\right) + \left(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}\right)$

    $\displaystyle > 1 + \frac{1}{2} + \left(\frac{1}{4} + \frac{1}{4}\right) + \left(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}\right) = 1 + 3\left(\frac{1}{2}\right)$

    And therefore, in general:

    $\displaystyle H_{2^k} \geq 1 + k\left(\frac{1}{2}\right)$.

    And since the subsequence $\displaystyle \{H_{2^k}\}$ is unbounded, the sequence $\displaystyle \{H_n\}$ diverges.

    And therefore it is sufficient to say that if the sequence diverges, no integer can be reached and therefore:

    $\displaystyle \{H_n\} \notin \mathbb{Z}$
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  5. #5
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    Quote Originally Posted by Aryth View Post
    There is a classical proof provided by Nicole Oresme, where you consider the subsequence $\displaystyle \{H_{2^k}\}^{\infty}_{k=0}$:

    $\displaystyle H_1 = 1 = 1 + 0\left(\frac{1}{2}\right)$

    $\displaystyle H_2 = 1 + \frac{1}{2} = 1 + 1\left(\frac{1}{2}\right)$

    $\displaystyle H_4 = 1 + \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{4}\right)$

    $\displaystyle > 1 + \frac{1}{2} + \left(\frac{1}{4} + \frac{1}{4}\right) = 1 + 2\left(\frac{1}{2}\right)$

    $\displaystyle H_8 = 1 + \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{4}\right) + \left(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}\right)$

    $\displaystyle > 1 + \frac{1}{2} + \left(\frac{1}{4} + \frac{1}{4}\right) + \left(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}\right) = 1 + 3\left(\frac{1}{2}\right)$

    And therefore, in general:

    $\displaystyle H_{2^k} \geq 1 + k\left(\frac{1}{2}\right)$.

    And since the subsequence $\displaystyle \{H_{2^k}\}$ is unbounded, the sequence $\displaystyle \{H_n\}$ diverges.

    And therefore it is sufficient to say that if the sequence diverges, no integer can be reached and therefore:

    $\displaystyle \{H_n\} \notin \mathbb{Z}$
    Ahem. I have to say I don't follow you there.

    You have shown that 1 + 1/2 + 1/3 + ... + 1/n can be made arbitrarily large by taking n large enough. But how do you know it can't be an integer?
    Last edited by ThePerfectHacker; Sep 13th 2008 at 05:16 PM. Reason: Question was double-posted
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  6. #6
    o_O
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    $\displaystyle S = 1 + 2 + 3 + \hdots + n \qquad n \in \mathbb{Z^{+}} \ \ \Rightarrow \ \ S \in \mathbb{Z^{+}}$

    The sequence $\displaystyle \{n\}$ diverges but you can't conclude S isn't an integer for all $\displaystyle n$.
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  7. #7
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    It is also here.
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