Show $\sum_{k=1}^{2n-1}\frac{\left(-1\right)^{k-1}k}{\binom{2n}{k}}=\frac{n}{n+1}$

\begin{align} \sum_{k=1}^{2n-1}\frac{(-1)^{k-1}k}{\binom{2n}{k}} &=(2n+1)\sum_{k=1}^{2n-1}\frac{(-1)^{k-1}k}{(2n+1)\binom{k+(2n-k)}{k}}\\ &=(2n+1)\sum_{k=1}^{2n-1}(-1)^{k-1}k\int_0^1 t^k (1-t)^{2n-k} \mathrm{d}t\\ &=(2n+1)\int_0^1 \sum_{k=1}^{2n-1}(-1)^{k-1}kt^k (1-t)^{2n-k} \mathrm{d}t\\ &=-(2n+1)\int_0^1 (t-1)^{2n}\sum_{k=1}^{2n-1}k\left(\frac{t}{t-1}\right)^k \mathrm{d}t\\ &=-(2n+1)\int_0^1 (t-1)^{2n}(t - 1) \left(t - \left(\frac{t}{t - 1}\right)^{2 n} (t - 2 n)\right) \mathrm{d}t\\ &=(2n+1)\int_0^1 (t - 1) (t^{2 n} (t - 2 n) - t (t-1)^{2 n}) \mathrm{d}t\\ &= (2n+1)\int_0^1 t(t - 1) (t^{2 n} - (t-1)^{2 n}) \mathrm{d}t\\ &\quad-2n(2n+1)\int_0^1 (t - 1) t^{2 n} \mathrm{d}t \\ &= 0-2n(2n+1)\left(\frac{1}{2n+2}-\frac{1}{2n+1}\right) \\ &=\frac{n}{n+1} \end{align}