Question : How can we prove the following equations combinatorially? $$\begin{eqnarray}\sum_{k=1}^{n}\frac{\binom{n-1}{k-1}}{\binom{n+k}{k}}&=&\frac 12\\\sum_{k=1}^{n}\frac{k\binom{n}{k}}{\binom{n+k}{k}}&=&\frac n2\end{eqnarray}$$
Recently I've known that the following equation holds for every $n\in\mathbb N$. $$\sum_{k=1}^{n}k\binom{2n}{n+k}=\frac n2\binom{2n}{n}$$
By the way, I noticed the following relations : $$\sum_{k=1}^{n}k\binom{2n}{n+k}=\frac n2\binom{2n}{n}\iff\sum_{k=1}^{n}\frac{\binom{n-1}{k-1}}{\binom{n+k}{k}}=\frac 12\iff \sum_{k=1}^{n}\frac{k\binom{n}{k}}{\binom{n+k}{k}}=\frac n2$$
I've already known that these equations are proved algebraically. Can anyone help?
I do not know if this counts as a combinatorial proof, but $$ \sum_{k=1}^{n}\frac{\binom{n-1}{k-1}}{\binom{n+k}{k}}=n\sum_{k=1}^{n}\binom{n-1}{k-1}\frac{\Gamma(k+1)\Gamma(n)}{\Gamma(n+k+1)}=n\int_{0}^{1}\sum_{k=1}^{n}\binom{n-1}{k-1}x^k(1-x)^{n-1}\,dx $$ hence: $$ \sum_{k=1}^{n}\frac{\binom{n-1}{k-1}}{\binom{n+k}{k}}=n\int_{0}^{1}x(1+x)^{n-1}(1-x)^{n-1}\,dx=\frac{1}{2}\int_{0}^{1}\frac{d}{dx}(1-x^2)^n\,dx=\color{red}{\frac{1}{2}}. $$
In probabilistic terms, we may consider a random path among all the paths in a $n \times n$ grid, starting in the bottom left corner and ending in the upper right corner, in which each step is a unit step towards north or east. The probability that the last step is towards north is obviously $\frac{1}{2}$ by symmetry, and our sum should be another way for computing such a probability by separating our paths into suitable equivalence classes, depending on the position of the last intersection with the diagonal.