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} $ 


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 $0$.
