combinatorics - Proving $sum_{k=1}^{n}binom{n-1}{k-1}{binom{n+k}{k}}^{-1}=frac 12$ combinatorially

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?