A proof for the identity $\sum_{n=r}^{\infty} {n \choose r}^{-1} = \frac{r}{r-1}$ 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
 A: $${n \choose r}^{-1}$$
$$=\frac{r!(n-r)!}{n!}$$
$$=\frac{r\Gamma(r) \Gamma (n-r+1)}{\Gamma (n+1)}$$
$$=rB(r,n-r+1)$$
$$=r \int_{0}^{1} x^{n-r} (1-x)^{r-1} dx$$
$$=r \int_{0}^{1} x^n x^{-r} (1-x)^{r-1} dx$$
Also note,
$$\sum_{n=r}^{\infty} x^n=\sum_{n=0}^{\infty} x^{n+r}=\frac{x^r}{1-x}$$
So that,
$$\sum_{n=r}^{\infty} {n \choose r}^{-1}$$
$$=r \int_{0}^{1} (1-x)^{r-2} dx$$
$$=\frac{r}{r-1}$$
A: \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}
A: This is basically a generalization of the approach from this answer:
\begin{align*}
\sum_{n=r}^\infty \frac1{\binom nr} 
&= r! \sum_{n=r}^\infty \frac1{n(n-1)\cdots(n-r+1)}\\
&= \frac{r!}{r-1} \sum_{n=r}^\infty \frac{n-(n-r+1)}{n(n-1)\cdots(n-r+1)}\\
&= \frac{r!}{r-1} \sum_{n=r}^\infty \left(\frac1{(n-1)\cdots(n-r+1)} - \frac1{n(n-1)\dots(n-r+2)}\right)\\
&\overset{(1)}= \frac{r!}{r-1} \cdot \frac1{(r-1)!} = \\
&= \frac{r}{r-1}
\end{align*}
The equation $(1)$ is based on the fact that we get a telescoping sum and with the exception of the first term, all remaining terms "cancel out".
The above works only for $r\ne1$. But it is clear that for $r=1$ we get the harmonic series $\sum\limits_{n=1}^\infty \frac1n$, which is divergent.

To explain this more clearly, we have a sum of the form $$\sum_{r=n}^\infty (a_n-a_{n+1})= (a_r-a_{r+1})+ (a_{r+1}-a_{r+2})+\dots.$$ Since $a_n\to 0$, we get that the sum is simply $a_r$, since the partial sum is $$\sum_{r=n}^N (a_n-a_{n+1}) = (a_r-a_{r+1})+(a_{r+1}-a_{r+2})+\dots+(a_N-a_{N+1}) = a_r-a_{N+1}.$$ In our case $a_n=1/[(n-1)\cdots(n-r)]$.

After some rewriting, we can see that this is a sum which has been calculated on this site a few times, namely $\sum_{k=1}^\infty \frac1{k(k+1)\cdots(k+s)}$. In our case, $s=r-1$. (Clearly, when I was trying to search for it before posting my answer, I did not choose the correct search queries.) A few posts about this sum found using the above search in Approach0:


*

*How do I evaluate this limit: $\lim_{n\to+\infty}\sum_{k=1}^{n} \frac{1}{k(k+1)\cdots(k+m)}$?

*How do we calculate this sum $\sum_{n=1}^{\infty} \frac{1}{n(n+1)\cdots(n+p)}$?

*Calculate the infinite sum $\sum_{k=1}^\infty \frac{1}{k(k+1)(k+2)\cdots (k+p)} $

*How to prove that $\sum\limits_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)...(n+k)} = \frac{1}{kk!}$ for every $k\geqslant1$
We can also find some posts about the finite sum, for example General formula for this sum $\sum_{k=1}^n\frac{1}{k(k+1)...(k+m)}$.
