Proving $$\sum_{n=0}^\infty \binom {m+n}{n}^{-1}=\frac{m}{m-1}$$
I'm not quite sure what to do with the sum of terms that combinations appear in denominators. I tried to solve the problem via splitting items, but failed.
But when I happened to see the following formula $$ \int_0^1 x^n (1-x)^m\,dx= \frac{n!m!}{(n+m+1)!} $$ I got some idea as follows
\begin{align} \sum_{n=0}^\infty \binom{m+n}{n}^{-1} &= \sum_{n=0}^\infty \, (m+n+1)\int_0^1 x^n (1-x)^m\,dx \\ &= \int_0^1 \left( \sum_{n=0}^\infty (m+n+1)x^n \right) (1-x)^m \,dx \\ &= \int_0^1 \left( \frac{m+1}{1-x}+\frac{x}{(1-x)^2} \right) (1-x)^m \,dx \\ &= (m+1)\int_0^1 (1-x)^{m-1} \,dx + \int_0^1 x (1-x)^{m-2} \,dx \\ &= \frac{m}{m-1} \end{align}
I am looking forward to other methods or a more deep-going idea to solve this problem such as solving it via the property of gamma function, I guess. Thanks for your help in advance!
