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Thread: Sum of inverses of primes

  1. #1
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    Sum of inverses of primes

    Assuming $\displaystyle p_n$ is the nth prime number, establish that the sum $\displaystyle 1/p_1 + 1/p_2 + ... + 1/p_n $ is never an integer.

    I don't know how to go about showing this. Any help is greatly appreciated, thanks!
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  2. #2
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    Let $\displaystyle S=1/p_1+1/p_2+\cdots+1/p_n$. This leads to $\displaystyle p_1p_2p_3\cdots p_nS=p_2p_3\cdots p_n+p_1p_3\cdots p_n+\cdots+p_1p_2\cdots p_{n-1}$.

    Assume to the contrary that $\displaystyle S $ is an integer; then $\displaystyle p_1 $ divides $\displaystyle p_2p_3\cdots p_n+p_1p_3\cdots p_n+\cdots+p_1p_2\cdots p_{n-1}$. Try to find a contradiction...

    By the way, using this idea can show that $\displaystyle 1/a_1+1/a_2+\cdots+1/a_n $ is not an integer if $\displaystyle (a_i,a_j)=1 $ for $\displaystyle i\neq j$, i.e., the $\displaystyle a_i $ are relatively prime in pairs.
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  3. #3
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    A different way, simpler maybe.

    I didn't want to resist the following solution. Again let $\displaystyle S=1/p_1+1/p_2+\cdots+1/p_n$; suppose $\displaystyle S $ is an integer.
    Each term in expasion of $\displaystyle p_2p_3\cdots p_nS$ is an integer except for $\displaystyle \displaystyle{\frac{p_2p_3\cdots p_n}{p_1}} $, but then $\displaystyle p_2p_3\cdots p_nS$ cannot be an integer.
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  4. #4
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    Quote Originally Posted by melese View Post
    I didn't want to resist the following solution. Again let $\displaystyle S=1/p_1+1/p_2+\cdots+1/p_n$; suppose $\displaystyle S $ is an integer.
    Each term in expasion of $\displaystyle p_2p_3\cdots p_nS$ is an integer except for $\displaystyle \displaystyle{\frac{p_2p_3\cdots p_n}{p_1}} $, but then $\displaystyle p_2p_3\cdots p_nS$ cannot be an integer.


    This is basically the very same nice solution you first posted, but with the conclusion shown in a slightly different way.

    Tonio
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