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Math Help - A sum hard to calculate

  1. #1
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    A sum hard to calculate

    Hi,

    I just can't think of a way to calculate this sum:



    Any help or a clue would be appriciated!
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by Krakowie View Post
    Hi,

    I just can't think of a way to calculate this sum:



    Any help or a clue would be appriciated!
    Do you have a typo? Do you mean

    \sum_{k=0}^{n}\binom{n}{k}\frac{k!}{(n+k+1)!}
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  3. #3
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    Yes, it's a typo, sorry.... I meant n choose k, i just couldn't get latex to work (
    Can you maybe solve it?
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  4. #4
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    \sum_{k=0}^n{n\choose k}\frac{k!}{(n+k+1)!} = \sum_{k=0}^n\frac{n!}{k!(n-k)!}\frac{k!}{(n+k+1)!} = \frac{n!}{(2n+1)!}\sum_{k=0}^n\frac{(2n+1)!}{(n-k)!(n+k+1)!}.

    Can you see why I inserted the (2n+1)! on the top and bottom of the fraction, and where to take it from there?
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  5. #5
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    Quote Originally Posted by Opalg View Post
    \sum_{k=0}^n{n\choose k}\frac{k!}{(n+k+1)!} = \sum_{k=0}^n\frac{n!}{k!(n-k)!}\frac{k!}{(n+k+1)!} = \frac{n!}{(2n+1)!}\sum_{k=0}^n\frac{(2n+1)!}{(n-k)!(n+k+1)!}.

    Can you see why I inserted the (2n+1)! on the top and bottom of the fraction, and where to take it from there?
    Thanks for your answer! I understand that now \frac{(2n+1)!}{(n-k)!(n+k+1)!} is a binomial {2n+1 \choose n-k} but I don't know what to do next... I've tried using the Vandermonde's identity for {k \choose k}{2n+1 \choose n-k} but it doesn't work since k is not a constant integer...Can you tell me what should I do now?
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  6. #6
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    \sum_{k=0}^n{2n+1 \choose n-k}=\sum_{k=0}^n{2n+1 \choose k}=\frac{1}{2}\sum_{k=0}^{2n+1}{2n+1 \choose k}
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  7. #7
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    Quote Originally Posted by emakarov View Post
    \sum_{k=0}^n{2n+1 \choose n-k}=\sum_{k=0}^n{2n+1 \choose k}=\frac{1}{2}\sum_{k=0}^{2n+1}{2n+1 \choose k}
    Thanks for the reply... How do you get that \sum_{k=0}^n{2n+1 \choose n-k}=\sum_{k=0}^n{2n+1 \choose k} ?
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  8. #8
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    \sum_{k=0}^n{2n+1 \choose n-k}=
    {2n+1 \choose n}+{2n+1 \choose n-1}+\dots+{2n+1 \choose 1}+{2n+1 \choose 0}=
    {2n+1 \choose 0}+{2n+1 \choose 1}+\dots+{2n+1 \choose n-1}+{2n+1 \choose n}=
    \sum_{k=0}^n{2n+1 \choose k}
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  9. #9
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    Or shift the index directly by replacing k with n-k:

    \begin{aligned}\displaystyle \sum_{0\le k \le n}\binom{2n+1}{n-k} & = \sum_{0\le n-k \le n}\binom{2n+1}{n-(n-k)} \\& = \sum_{-n\le -k \le 0}\binom{2n+1}{n-n+k} \\& = \sum_{0\le k \le n}\binom{2n+1}{k}. \end{aligned}
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  10. #10
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    Thank you so much for your help!
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