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Math Help - 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 <br />
p_n <br />
be the greatest prime not exceeding n.

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

    Now suppose <br />
{1 + \tfrac{1}<br />
{2} + ... + \tfrac{1}<br />
{{p_n }} + ... + \tfrac{1}<br />
{n}} \in <br />
\mathbb{Z}^ +  <br /> <br />
then a=<br />
\tfrac{{n!}}<br />
{{p_n }} \cdot \left( {1 + \tfrac{1}<br />
{2} + ... + \tfrac{1}<br />
{{p_n }} + ... + \tfrac{1}<br />
{n}} \right) \in \mathbb{Z}^ +  <br />
since <br />
\tfrac{{n!}}<br />
{{p_n }}<br />
is an integer

    <br />
a = \underbrace {\tfrac{{n!}}<br />
{{p_n }} \cdot \tfrac{1}<br />
{1} + \tfrac{{n!}}<br />
{{p_n }} \cdot \tfrac{1}<br />
{2} + ...}_{ \in \mathbb{Z}} + \tfrac{{n!}}<br />
{{p_n }} \cdot \tfrac{1}<br />
{{p_n }} + \underbrace {... + \tfrac{{n!}}<br />
{{p_n }} \cdot \tfrac{1}<br />
{n}}_{ \in \mathbb{Z}}<br />

    By (1) we have that <br />
p_n ^2 <br />
doesn't divide n! so 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 \{H_{2^k}\}^{\infty}_{k=0}:

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

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

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

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

    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)

     > 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:

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

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

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

    \{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 \{H_{2^k}\}^{\infty}_{k=0}:

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

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

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

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

    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)

     > 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:

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

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

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

    \{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; September 13th 2008 at 05:16 PM. Reason: Question was double-posted
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  6. #6
    o_O
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    S = 1 + 2 + 3 + \hdots + n \qquad n \in \mathbb{Z^{+}} \ \ \Rightarrow \ \ S \in \mathbb{Z^{+}}

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