How to prove combinatoric identity How would you prove this identity using combinatorics? Any hints or advice?
For all positive integers $n>1$,
$\sum_{k=0}^{n} \frac{1}{k+1} {n\choose k} (-1)^{k+1}=\frac{-1}{n+1} $ 
 A: We have:  $f(x) = \dfrac{1}{k+1}\cdot \displaystyle \sum_{k=0}^n \binom{n}{k}x^{k+1}\implies f'(x) = (1+x)^n\implies f(x) = f(0) + \displaystyle \int_{0}^x f'(t)dt = 0 + \displaystyle \int_{0}^x(1+t)^ndt = \displaystyle \int_{1}^{1+x}u^ndu= \dfrac{u^{n+1}}{n+1}|_{u=1}^{u=x+1}= \dfrac{(x+1)^{n+1}}{n+1}- \dfrac{1}{n+1}\implies f(-1) = -\dfrac{1}{n+1}$ . 
A: 
We obtain
  \begin{align*}
\color{blue}{\sum_{k=0}^n\frac{1}{k+1}\binom{n}{k}(-1)^{k+1}}
&=\frac{1}{n+1}\sum_{k=0}^n\binom{n+1}{k+1}(-1)^{k+1}\tag{1}\\
&=\frac{1}{n+1}\sum_{k=1}^{n+1}\binom{n+1}{k}(-1)^{k}\tag{2}\\
&=\frac{1}{n+1}(1-1)^{n+1}-\frac{1}{n+1}\tag{3}\\
&\color{blue}{=-\frac{1}{n+1}}
\end{align*}

Comment:


*

*In (1) we use the binomial identity $\binom{n+1}{k+1}=\frac{n+1}{k+1}\binom{n}{k}$.

*In (2) we shift the index to start from $k=1$.

*In (3) we apply the binomial summation formula and subtract $\frac{1}{n+1}$ as compensation for the lower limit starting with $1$ instead of $0$.
A: HINT 1: Notice that
$$\frac{1}{k+1}\binom{n}{k}=\frac{1}{n+1}\binom{n+1}{k+1}$$
HINT 2: Do you know the binomial theorem?
