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
$$
 \sum\limits_{k=1}^n \frac{{n \choose k}.{(-1)^{k}}}{k+1}
$$
 A: $\sum_{k=0}^n {n \choose k}x^k=(1+x)^n$
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 $\sum_{k=1}^n {n \choose k}\frac{(-1)^{k}}{k+1}=\frac{1}{n+1}-{n \choose 0}\frac{(-1)^0}{1}=\frac{1}{n+1}-1$
A: You can write $\binom nk \over k+1$ as $\binom {n+1}{k+1}\over n+1$ and also write $(-1)^k$ as $-(-1)^{k+1}$. Then the summation reduces to $$\frac {-1}{n+1}\sum_{k=1}^n\binom {n+1}{k+1}(-1)^{k+1}$$Now by changing the variable $k+1=i$ and using the expansion of $(1+x)^n$ you can easily get the answer. I leave the rest part of the solution to you. 
A: Start with: $\displaystyle \int_{0}^1x(1+x)^ndx = \displaystyle \int_{0}^1\displaystyle \sum_{k=1}^n \binom{n}{k}x^{k+1}dx$, and plug $x = -1$ into the new equation.
A: I am quite sure it is a duplicate, but in any case:
$$\sum_{k=1}^{n}\binom{n}{k}\frac{(-1)^k}{k+1}=\int_{0}^{1}\sum_{k=1}^{n}\binom{n}{k}(-x)^k\,dx=\int_{0}^{1}-1+(1-x)^n\,dx=\color{red}{-1+\frac{1}{1+n}}. $$
