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Math Help - Harmonic Numbers

  1. #1
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    Harmonic Numbers

    Challange Problem; for those who have never seen it, try this by yourselves.

    Prove that for n>1 , the sum  \displaystyle{1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}} is not an integer.

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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Maybe a solution: (first thing that came to my head...)

    1+1/2+...+1/n=H_n

    Hence:

    H_n - H_{n-1}=1/n

    n(H_n - H_{n-1})=1

    Suppose H_n integer, hence H_n - H_{n-1} not integer! BUT n(H_n - H_{n-1}) is integer (1) ==>H_n can't be integer.
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  3. #3
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Maybe a solution: (first thing that came to my head...)

    1+1/2+...+1/n=H_n

    Hence:

    H_n - H_{n-1}=1/n

    n(H_n - H_{n-1})=1

    Suppose H_n integer, hence H_n - H_{n-1} not integer! BUT n(H_n - H_{n-1}) is integer (1) ==>H_n can't be integer.
    Hi Also sprach Zarathustra, you're are saying essentially that the product of an integer and a rational (that is non-integer) cannot be an integer.
    To look at differently H_n-H_{n-1} is indeed not an integer ( 1/n), and also n(H_n-H_{n-1} )=n\cdot\frac{1}{n}=1.
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  4. #4
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Maybe a solution: (first thing that came to my head...)

    1+1/2+...+1/n=H_n

    Hence:

    H_n - H_{n-1}=1/n

    n(H_n - H_{n-1})=1

    Suppose H_n integer, hence H_n - H_{n-1} not integer! BUT n(H_n - H_{n-1}) is integer (1) ==>H_n can't be integer.


    Uuh?? We don't need the preceeding part to "...hence H_n-H_{n-1} not integer" to conclude this: it must be obvious that it is not an integer since for 1<n\in\mathbb{N}\,,\,\,0<\frac{1}{n}<1 ...and

    thus your conclusion doesn't follow: for example,  \frac{1}{3}\,,\,\frac{2}{3} aren't integers, but 3\cdot\frac{1}{3}\,,\,3\cdot\frac{2}{3} are.

    The proofs I know of this aren't hard but can be tricky. Hints for the most elementary one (that I know, of course):

    If H_n\in\mathbb{N}\,,\,\,then\,\,\,H_n\geq 2 . Choose now k\in\mathbb{N}\,\,\,s.t. \,\,\,1<2^k\leq n \Longrightarrow l.c.m. (1,2,\ldots,n)=2^k\cdot 3^s\cdot5^r\cdot\ldots , and now divide and sum and stuff.

    Tonio
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    Well for 1/1 + 1/2 + 1/4 +1/8 converges to two. 1/3 + 1/9 + 1/27 converges as well. This problem is really just asking if the series 1/p converges. P = prime

    I am sure that this harmonic series diverges, but I really don't know how to prove it. And if it diverges, it does not converge on an integer.
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