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Thread: finite alternating harmonic series

  1. #1
    MHF Contributor
    Dec 2012
    Athens, OH, USA

    finite alternating harmonic series

    This is similar to an existing thread, but sufficiently different, I think.
    For any positive integer n, let $\displaystyle S(n)=\Sigma^n_{k=1}\left({1\over2k-1}-{1\over2k}\right)={u_n\over d_n}$ with $\displaystyle u_n$ and $\displaystyle d_n$ relatively prime. The question concerns $\displaystyle u_n$ and $\displaystyle d_n$.

    One result. The exact power of 2 that divides $\displaystyle d_n$ is floor(log2(n))+1.

    Here's the proof:
    finite alternating harmonic series-mhfalternatingharmonic.png

    What else can one say about the sequence $\displaystyle d_n$? The exact power of 3 that divides $\displaystyle d_n$ seems to be an increasing sequence, but I can't find any closed formula as for 2. The exact power of 5 that divides $\displaystyle d_n$ is not even an increasing sequence.

    The integers $\displaystyle u_n$ and $\displaystyle d_n$ get quite large. I know almost nothing about the $\displaystyle u_n$.
    $\displaystyle u_2$, $\displaystyle u_3$, $\displaystyle u_5$, $\displaystyle u_8$ and $\displaystyle u_9$ are prime.

    Java has a pretty sophisticated probalistic prime testing algorithm. Java says with probability greater than 1-one millionth that $\displaystyle u_{254}$ is prime; $\displaystyle u_{254}$ has 221 decimal digits.

    (I inadvertenly attached a thumbnail and don't know how to delete it; ignore it.)
    Attached Thumbnails Attached Thumbnails finite alternating harmonic series-alternatingseries.png  
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