Proving that $\sum_{k=0}^{n}\frac{(-1)^k}{{n\choose k}}=[1+(-1)^n] \frac{n+1}{n+2}.$ Various examinations ask students to prove that $$\frac{1}{{n \choose k}}=(n+1) \int_{0}^{1} x^k (1-x)^{n-k} dx ~~~~~(1)$$ by evaluating  the integral $\int_{0}^{1}
(tx+1-x)^n~dt$ two ways. When I came across (1), I could prove that $$\sum_{k=0}^{n}\frac{(-1)^k}{{n\choose k}}=[1+(-1)^n] \frac{n+1}{n+2}.~~~~(2)$$ as below:  $$S_n=\frac{1}{n+1}\sum_{k=0}^{n}\frac{(-1)^k}{{n\choose k}}=\int_{0}^1 \sum_{k=0}^{n} (-1)^k  x^k (1-x)^{n-k} dx =  \int_{0}^{1}  (1-x)^n \sum_{k=0}^{n}\left(\frac{-x}{1-x} \right)^k dx =\int_{0}^{1}(1-x)^n \frac{ \left( \frac{-x}{1-x}\right)^{n+1}-1}{\left(\frac{-x}{1-x}\right)-1}dx$$
$$=\int_{0}^{1}[(1-x)^{n+1}-(-x)^{n+1}] dx =\frac{1+(-1)^n}{n+2}.~~~~(3)$$
Therefore, $$S_{2m+1}=0 ~~~~(4) ~~~ \mbox{and}~~~ S_{2m}= \frac{2m+1}{m+1}.~~~~(5)$$
Now the question is: Can one prove (2) in some other way(s)? 

The general formula is: 

$$\sum\limits_{k=0}^n\frac{(-1)^k}{\binom x k} = \left(1+\frac{(-1)^n}{\binom {x+1} {n+1}}\right)\frac{x+1}{x+2}$$

Value range: $\enspace n\in\mathbb{N}_0~,~~ x\in\mathbb{C}\setminus\{n-k|k\in\mathbb{N}\}$ 
The proof by induction with respect to $~n~$ is based on the following equation:
$$\frac{1}{\binom x {n+1}}\frac{x+2}{x+1} = \frac{1}{\binom {x+1} {n+1}} + \frac{1}{\binom {x+1} {n+2}}$$ 
