Combinatorial identity of $\sum\limits_{k=1}^n \frac{{n \choose k}.{(-1)^{k}}}{k+1}$ I don't know how to deal with this example: Find a closed form of
$$
 \sum\limits_{k=1}^n \frac{{n \choose k}.{(-1)^{k}}}{k+1}
$$

$\sum_{k=0}^n {n \choose k}x^k=(1+x)^n$
Integrating both sides to get
$\sum_{k=0}^n {n \choose k}\frac{x^{k+1}}{k+1}=\frac{(1+x)^{n+1}}{n+1}+c$
For $x=0$ we have $0=\frac{1}{n+1}+c$, i.e. $c=-\frac{1}{n+1}$
$x\sum_{k=0}^n {n \choose k}\frac{x^{k}}{k+1}=\frac{(1+x)^{n+1}}{n+1}-\frac{1}{n+1}$
For $x=-1$ we have
$-\sum_{k=0}^n {n \choose k}\frac{(-1)^{k}}{k+1}=\frac{(1-1)^{n+1}}{n+1}-\frac{1}{n+1}=-\frac{1}{n+1}$
So $\sum_{k=1}^n {n \choose k}\frac{(-1)^{k}}{k+1}=\frac{1}{n+1}-{n \choose 0}\frac{(-1)^0}{1}=\frac{1}{n+1}-1$
