Prove $\sum\limits_{k=0}^n \binom{n}{k} \frac{(-1)^k}{k+1} = \frac{1}{n+1}$.

Question

I managed to get
$$\sum_{k=0}^n \binom{n+1}{k+1}(-1)^k$$ on the left side, but I don't know how to proceed from here. thanks in advance.

Answer 1

Expand $(1-1)^{n+1}$ by using the Binomial Theorem: $$0=(1-1)^{n+1}=1-\sum_{k=0}^{n}\binom{n+1}{k+1}(-1)^{k}=1-(n+1)\sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^{k}}{k+1}.$$

Answer 2

$$\begin{eqnarray*}\sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^k}{k+1}&=&\int_{0}^{1}\sum_{k=0}^{n}\binom{n}{k}(-x)^k\,dx\\&=&\int_{0}^{1}(1-x)^n\,dx\\&=&\int_{0}^{1}z^n\,dz = \color{red}{\frac{1}{n+1}}.\end{eqnarray*}$$