Understanding steps in a proof of $\sum_{k=1}^n\binom{n-1}{k-1}\binom{2n-1}{k}^{-1}=\frac{2}{n+1}$ So, the task is to prove that:
$$\sum_{k=1}^n\binom{n-1}{k-1}\binom{2n-1}{k}^{-1}=\frac{2}{n+1}$$
I tried different methods but none led me to the solution. I looked up the solution and I can't even understand what is done there. Could you please explain me what is done in each step (except the first one):
$$\sum_{k=1}^n\binom{n-1}{k-1}\binom{2n-1}{k}^{-1}=\frac{(n-1)!}{(2n-1)!}\sum_{k=1}^n\frac{k(2n-k-1)!}{(n-k)!}$$
$$=\frac{2n}{2n-1}\binom{2n-2}{n-1}^{-1}\sum_{k=1}^n\binom{2n-k-1}{n-1}-\binom{2n-1}{n-1}^{-1}\sum_{k=1}^n\binom{2n-k}{n}$$
$$=\frac{2n}{2n-1}\binom{2n-2}{n-1}^{-1}\binom{2n-1}{n-1}-\binom{2n-1}{n-1}^{-1}\binom{2n}{n+1}=\frac{2}{n+1}$$
 A: $$\begin{align}&\sum_{k=1}^{n}\binom{n-1}{k-1}\binom{2n-1}{k}^{-1}\\&=\sum_{k=1}^{n}\frac{(n-1)!}{(k-1)!(n-k)!}\cdot\frac{k!(2n-k-1)!}{(2n-1)!}\\&=\frac{(n-1)!}{(2n-1)!}\sum_{k=1}^{n}\frac{\color{red}{k}(2n-k-1)!}{(n-k)!}\\&=\frac{(n-1)!}{(2n-1)!}\sum_{k=1}^{n}\frac{(\color{red}{2n-(2n-k)})(2n-k-1)!}{(n-k)!}\\&=\frac{(n-1)!}{(2n-1)!}\sum_{k=1}^{n}\left(\frac{2n(2n-k-1)!}{(n-k)!}-\frac{(2n-k)(2n-k-1)!}{(n-k)!}\right)\\&=\frac{(n-1)!}{(2n-1)!}\sum_{k=1}^{n}\frac{2n(2n-k-1)!}{(n-k)!}-\frac{(n-1)!}{(2n-1)!}\sum_{k=1}^{n}\frac{(2n-k)(2n-k-1)!}{(n-k)!}\\&=\frac{2n}{2n-1}\cdot\frac{(n-1)!(n-1)!}{(2n-2)!}\sum_{k=1}^{n}\frac{(2n-k-1)!}{(n-1)!(n-k)!}-\frac{(n-1)!n!}{(2n-1)!}\sum_{k=1}^{n}\frac{(2n-k)!}{(n-k)!n!}\\&=\frac{2n}{2n-1}\binom{2n-2}{n-1}^{-1}\color{blue}{\sum_{k=1}^{n}\binom{2n-k-1}{n-1}}-\binom{2n-1}{n-1}^{-1}\color{green}{\sum_{k=1}^{n}\binom{2n-k}{n}}\\&=\frac{2n}{2n-1}\binom{2n-2}{n-1}^{-1}\color{blue}{\binom{2n-1}{n-1}}-\binom{2n-1}{n-1}^{-1}\color{green}{\binom{2n}{n+1}}\\&=\frac{2}{n+1}\end{align}$$
A: In the given approach, you are writing your sum as a telescopic one.
Using Euler's beta function:
$$\begin{eqnarray*}\sum_{k=1}^{n}\binom{n-1}{k-1}\binom{2n-1}{k}^{-1} &=& \sum_{k=1}^{n}\binom{n-1}{k-1}\frac{k!(2n-k-1)!}{(2n-1)!}\\&=&\sum_{k=1}^{n}\binom{n-1}{k-1}\frac{\Gamma(k+1)\Gamma(2n-k)}{\Gamma(2n)}\\&=&\sum_{k=1}^{n}\binom{n-1}{k-1}k\int_{0}^{1}x^{k-1}(1-x)^{2n-k-1}\,dx\\&=&\int_{0}^{1}(1-x)^{n-1}\left(1+x(n-1)\right)\,dx\\&=&\int_{0}^{1}(n+x-nx)x^{n-1}\,dx\\&=&\frac{n}{n}+\frac{1}{n+1}-\frac{n}{n+1}=\color{red}{\frac{2}{n+1}}.\end{eqnarray*}$$
