TITLE: A proof for the identity $\sum_{n=r}^{\infty} {n \choose r}^{-1} = \frac{r}{r-1}$
QUESTION [15 upvotes]: Do you have any idea to prove the following identity via a combinatorial (or algebraic) method? 
$\sum_{n=r}^{n=\infty} {n \choose r}^{-1} = \frac{r}{r-1}$
This is Exercise 71 in Chapter 2 of the book Chen C.C., Koh K.M. Principles and techniques in combinatorics. The book does not give a solution, although it mentions: "see H. W. Gould, Combinatorial Identities, Morgantown, W.V. (1972), 18-19".
many thanks in advance,
Shahram

REPLY [3 votes]: \begin{align}
\sum_{n=r}^\infty \binom{n}{r}^{-1}
&=
\sum_{n=r}^\infty \, (n+1)\int_0^1 x^{n-r} (1-x)^{r}\,dx
\\ &=
\int_0^1 \left( \sum_{n=r}^\infty (n+1)x^{n-r} \right) (1-x)^r \,dx
\\ &=
\int_0^1 \frac{1+r-rx}{(1-x)^2}  (1-x)^r \,dx
\\ &=
\frac{r}{r-1}
\end{align}