TITLE: True? $\sum_{n=1}^\infty {n+m-1 \choose m}^{-1} = 1 + \frac {1}{m-1} $
QUESTION [6 upvotes]: 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} $

REPLY [11 votes]: 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*}