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Prove (or disprove) that for all positive integers $m>1$,

$ \displaystyle \sum_{n=1}^\infty {n+m-1 \choose m}^{-1} = 1 + \frac {1}{m-1} $

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1 Answer 1

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Using the formula ${1/ {n\choose r}}=(n+1)\int_0^1 u^r(1-u)^{n-r}\,du,$ gives: $${1\over {n+m-1 \choose m}}=\int_0^1 (n+m)\,u^m\,(1-u)^{n-1}\,du. $$ Thus, adding over $n$ gives \begin{eqnarray*} \sum_{n=1}^\infty {1\over {n+m-1 \choose m} } &=&\int_0^1 \sum_{n=1}^\infty (n+m)\,u^m\,(1-u)^{n-1}\,du\\[8pt] &=&\int_0^1 (1+mu)\,u^{m-2} \,du\\[8pt] &=&{1\over m-1}+{m\over m}\\[8pt] &=&{1\over m-1}+1. \end{eqnarray*}

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  • $\begingroup$ Maybe the additional "brackets" in my question wasn't necessary. I've changed it. Sorry for the inconvenience. $\endgroup$
    – GohP.iHan
    May 20, 2014 at 23:06
  • $\begingroup$ Whoops! Time to recalculate. $\endgroup$
    – user940
    May 20, 2014 at 23:07

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