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

Question

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.

Answer 1

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 $$\sum_{k=0}^{2n}\frac{(-1)^k}{{2n \choose k}}=\frac{2n+1}{n+1} \implies 1-\frac{S_n}{n}+1=\frac{2n+1}{n+1} \implies S_n=\frac{n}{n+1}.$$

Answer 2

\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}

Answer 3

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.