Alternate complex binomial series sum 
Calculation of $\displaystyle \sum^{2n-1}_{r=1}(-1)^{r-1}\cdot r\cdot \frac{1}{\binom{2n}{r}}$ is

My Try: Using $$\int^{1}_{0}x^m(1-x)^ndx = \frac{1}{(m+n+1)}\cdot \frac{1}{\binom{m+n}{n}}$$
So $\displaystyle \int^{1}x^{2n-r}(1-x)^r=\frac{1}{2n}\cdot \frac{1}{\binom{2n}{r}}$
Sum convert into $\displaystyle 2n\sum^{2n-1}_{r=1}(-1)^{r-1}r\int^{1}_{0}x^{2n-r}(1-x)^rdx$ 
$\displaystyle \Longrightarrow 2n \int^{1}_{0}x^{2n}\sum^{2n-1}_{r=1}(-1)^{r-1}\cdot r \cdot \bigg(1-\frac{1}{x}\bigg)^rdx$
Could some help me to solve it , Thanks 
 A: $\text{Lemma: } \sum_{k=0}^{n-1} k x^k=\frac{(n-1)x^n+1}{(x-1)}-\frac{(x^n-1)}{(x-1)^2}
$$\text{Proof: }\sum_{k=0}^{n-1} x^k=\frac{x^n-1}{x-1}\implies\sum_{k=0}^{n-1} k x^{k-1}=\frac{n(x-1)x^{n-1}-(x^n-1)}{(x-1)^2}$
$$\sum^{2n-1}_{r=1}(-1)^{r-1}\cdot r\cdot \frac{1}{\binom{2n}{r}}=\sum^{2n-1}_{r=1}(-1)^{r-1}\cdot r\cdot(2n+1)\int_0^1t^{2n-r}(1-t)^{r}dt$$$$=-(2n+1)\int_0^1t^{2n}\sum_{r=1}^{2n-1}r\Big(1-\frac{1}{t}\Big)^rdt$$$$=-(2n+1)\int_0^1t^{2n}\bigg(\frac{(2n-1){\Big(1-\frac{1}{t}\Big)}^{2n}+1}{\frac{-1}{t}}-\frac{{\Big(1-\frac{1}{t}\Big)}^{2n}-1}{(\frac{-1}{t})^2}\bigg)dt$$
$$=(2n+1)\int_0^1\bigg({(2n-1)t{(1-t)}^{2n}+t^{2n+1}}+t^2{(1-t)}^{2n}-t^{2n+2}\bigg)dt$$
$$=(2n+1)\bigg(\frac{(2n-1)}{(2n+2)(2n+1)}+\frac{1}{2n+2}+\frac{2}{(2n+1)(2n+2)(2n+3)}-\frac{1}{2n+3}\bigg)=\frac{n}{n+1}$$
$\blacksquare$
A: Here is a more elementary method to find the sum which gives a more general result. Let $\displaystyle f(n)=\sum^{n-1}_{r=1}(-1)^{r-1}\frac{r}{\binom{n}{r}}$. We have been asked to find $\displaystyle f(2n)$. Note that $\displaystyle \frac{1}{\binom{n+1}{k+1}}+\frac{1}{\binom{n+1}{k}}=\frac{n+2}{n+1}\frac{1}{\binom{n}{k}}$. We will use this identity twice to telescope the sum.
$$f(n)=\sum^{n-1}_{r=0}(-1)^{r-1}\frac{r}{\binom{n}{r}}=\frac{n+1}{n+2}\sum^{n-1}_{r=0}\bigg((-1)^{r-1}\frac{r}{\binom{n+1}{r}}-(-1)^{r}\frac{r}{\binom{n+1}{r+1}}\bigg)$$$$=\frac{n+1}{n+2}\sum^{n-1}_{r=0}\bigg((-1)^{r-1}\frac{r}{\binom{n+1}{r}}-(-1)^{r}\frac{r+1}{\binom{n+1}{r+1}}+\frac{(-1)^r}{\binom{n+1}{r+1}}\bigg)=\frac{n+1}{n+2}\bigg((-1)^n\frac{n}{n+1}+\sum^{n}_{r=1}\frac{(-1)^{r-1}}{\binom{n+1}{r}}\bigg)$$$$=\frac{n+1}{n+2}\Bigg((-1)^n\frac{n}{n+1}+\frac{n+2}{n+3}\sum^{n}_{r=1}\bigg(\frac{(-1)^{r-1}}{\binom{n+2}{r}}-\frac{(-1)^r}{\binom{n+2}{r+1}}\bigg)\Bigg)=\frac{n+1}{n+2}\Bigg((-1)^n\frac{n}{n+1}+\frac{1-(-1)^n}{n+3}\Bigg)$$
$$\implies f(n)=\frac{n}{n+2}(-1)^n+\frac{n+1}{(n+2)(n+3)}\big(1-(-1)^n\big)=\left \{
  \begin{aligned}
    &\ \ \ \ \ \frac{n}{n+2}, && \text{if}\ n \text{ is even} \\
    &-\frac{n-1}{n+3}, && \text{if } n \text{ is odd}
  \end{aligned} \right.$$
$$\therefore f(2k+1)=-\frac{k}{k+2}\text{ and }f(2k)=\frac{k}{k+1}\text{, as desired.}$$
$\blacksquare$
