Show $\sum_{k=1}^{2n-1}\frac{\left(-1\right)^{k-1}k}{\binom{2n}{k}}=\frac{n}{n+1}$ How it can be shown that:

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

for $1 \le n$

I tried to use this method , but that was not helpful, Also I tried to use the following identity:
$$\frac{1}{\binom{2n+1}{k}}+\frac{1}{\binom{2n+1}{k+1}}=\frac{2n+2}{2n+1}\ \frac{1}{ \binom{2n}{k}}$$
and use some telescoping property , but again that did not help me.
Please if it's possible,then do the proof using elementary ways.
 A: \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}
A: Another way
Let $$S_n=\sum_{k=0}^{2n-1}\frac{(-1)^{k-1} k}{{2n\choose k}}$$
Let us change $k$ to $2n-k$ (symmetry property of summation) in above to have
$$S_n=\sum_{k=1}^{2n-1} \frac{(-1)^{2n-k-1} (2n-k)}{{2n \choose 2n-k}}=2n \sum_{k=1}^{2n-1} \frac{(-1)^{k-1}}{{2n \choose k}}-\sum_{k=1}^{2n-1} \frac{(-1)^{k-1}k}{{2n\choose k}}$$
$$\implies S_n+S_n=2n \sum_{k=1}^{n} \frac{(-1)^{k-1}}{{2n \choose k}}\implies S_n=n \sum_{k=1}^{2n-1} \frac{(-1)^k}{{2n \choose k}}$$
Next, we use an interesting result at MSE:
Proving that $\sum_{k=0}^{n}\frac{(-1)^k}{{n\choose k}}=[1+(-1)^n] \frac{n+1}{n+2}.$
In this we change $n$ to $2n$ to get
$$\sum_{k=0}^{2n}\frac{(-1)^k}{{2n \choose k}}=\frac{2n+1}{n+1} \implies 1-\frac{S_n}{n}+1=\frac{2n+1}{n+1} \implies S_n=\frac{n}{n+1}.$$
A: Perhaps we could use the following formula to get rid of the binomial coefficient at the denominator : $$ \left(\forall k\in\mathbb{N}\right),\ \displaystyle\frac{\left(-1\right)^{k-1}}{\binom{2n}{k}}=k\displaystyle\sum_{k=0}^{k-1}{\displaystyle\frac{\left(-1\right)^{i}}{2n-i}\displaystyle\binom{k-1}{i}} $$
I managed to find the result using it, but the calculations were too long.
