The conclusion of # 4307 states $$\frac{2n+2}{2n+1}\binom{2n}{k}^{-1} = \binom{2n+1}{k}^{-1} +\binom{2n+1}{k+1}^{-1}$$

Therefore, $$\begin{align*} &\frac{2n+2}{2n+1}\sum_{k=1}^{2n-1}(-1)^{k-1}\binom{2n}{k}^{-1} \\=\ &\sum_{k=1}^{2n-1}(-1)^{k-1}\left(\binom{2n+1}{k}^{-1}+\binom{2n+1}{k+1}^{-1}\right) \\ =\ &\left(\binom{2n+1}{1}^{-1} + \binom{2n+1}{2}^{-1} \right) \\ -&\left(\binom{2n+1}{2}^{-1} + \binom{2n+1}{3}^{-1} \right) \\ &\cdots \\ +&\left(\binom{2n+1}{2n-1}^{-1} + \binom{2n+1}{2n}^{-1} \right) \\=\ &\binom{2n+1}{1}^{-1} + \binom{2n+1}{2n}^{-1} \\ =\ &\frac{1}{2n+1} + \frac{1}{2n+1} \\=\ &\frac{2}{2n+1} \end{align*}$$

$$\therefore\quad \sum_{k=1}^{2n-1}(-1)^{k-1}\binom{2n}{k}^{-1} = \frac{2}{2n+1}\cdot\frac{2n+1}{2n+2}=\boxed{\frac{1}{n+1}}$$