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Math Help - Prove this is not an integer

  1. #1
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    Prove this is not an integer

    Let p_{i} \ \ \ \ \ 1 \leq i \leq n be from the first to the nth prime numbers, prove that  \frac {1}{p_{1}} + \frac {1}{p_{2}} + . . . + \frac {1}{p_{n}} is never an integer.

    Proof.

    I use induction, but I don't know if it is okay.

    The first prime number is 2, 1/2 is not an integer, so 1 is okay.

    Let  \sum ^{k}_{1} \frac {1}{p_{i}} be not an integer.

    Then  \sum ^{k+1}_{1} \frac {1}{p_{i}} = \sum ^{k}_{1} \frac{1}{p_{i}} + \frac {1}{p_{k+1}} is also not an integer.
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Let p_{i} \ \ \ \ \ 1 \leq i \leq n be from the first to the nth prime numbers, prove that  \frac {1}{p_{1}} + \frac {1}{p_{2}} + . . . + \frac {1}{p_{n}} is never an integer.
    Note,
    \frac{1}{p_1}+...+\frac{1}{p_n} = \frac{\sum_{k=1}^n \prod_{i=1,i\not =k}^n p_i}{\prod_{i=1}^n p_i}.
    If that is an integer then if we multiply it by \prod_{i=1}^{n-1}p_i it will stay an integer.
    If we do that we are left with,
    \frac{\sum_{k=1}^n \prod_{i=1,i\not =k}^n p_i}{p_n}.
    If we expand the summation there will be n products each product will have a p_n factor except the last one (when k=n). Thus, we have the first n-1 terms divisible by p_n but not the last term, hence this cannot be an integer.
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