Proving that $\sum_{k=0}^{n}\frac{(-1)^k}{{n\choose k}}=[1+(-1)^n] \frac{n+1}{n+2}.$

A telescoping approach:

We obtain \begin{align*} \color{blue}{\sum_{k=0}^n\frac{(-1)^k}{\binom{n}{k}}} &=\sum_{k=0}^n(-1)^k\frac{k!(n-k)!}{n!}\\ &=\sum_{k=0}^n(-1)^k\frac{k!(n-k)!}{n!}\cdot\frac{(n-(k-1))+(k+1)}{n+2}\\ &=\frac{n+1}{n+2}\sum_{k=0}^n(-1)^k\frac{k!(n-(k-1))!+(k+1)!(n-k)!}{(n+1)!}\\ &=\frac{n+1}{n+2}\sum_{k=0}^n\left((-1)^k\frac{1}{\binom{n+1}{k}}-(-1)^{k+1}\frac{1}{\binom{n+1}{k+1}}\right)\\ &\,\,\color{blue}{=\frac{n+1}{n+2}\left(1+(-1)^n\right)} \end{align*}

and the claim follows.

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: $$\frac{1}{\binom x {n+1}}\frac{x+2}{x+1} = \frac{1}{\binom {x+1} {n+1}} + \frac{1}{\binom {x+1} {n+2}}$$

I think that the intuition about your result can be the following:

$f(n,k):=\int_0^1 x^k(1-x)^{n-k}>0$

and you can observe that

$1=1^n=((1-x)+x)^n=\sum_{k=0}^n\binom{n}{k}(1-x)^{n-k}x^k$

so if you integrate the two members with respect to $x$ you get that

$\int_0^11dx=1=$

$\int_0^1 (\sum_{k=0}^n\binom{n}{k}(1-x)^{n-k}x^k)dx=$

$=\sum_{k=0}^n\binom{n}{k}\int_0^1x^n(1-x)^{n-k}dx$

so

$1=\sum_{k=0}^n\binom{n}{k}f(n,k)$

and you can observe that the identity is satisfied when

$f(n,k)=\frac{1}{\binom{n}{k}}\frac{1}{n+1}$

So we want prove by induction on $k>1$ for all fixed $n\geq k$ that $f(n,k)=\frac{1}{\binom{n}{k}}\frac{1}{n+1}$

For $k=0$ you have that

$f(n,0)=\int_0^1x^0(1-x)^{n-0}dx=-\frac{1}{n+1}[(1-x)^{n+1}]|_0^1=\frac{1}{n+1}$

Now we can hypothesize that the sentence is true for some $k-1$ and we want prove that it is true for $k$:

$f(n,k)=\int_0^1x^k(1-x)^{n-k}dx=$

$-\frac{1}{n-k+1}\int_0^1x^kD((1-x)^{n-k+1})dx=$

$=\frac{1}{n-k+1}k\int_0^1x^{k-1}(1-x)^{n-(k-1)}dx=$

$=\frac{k}{n-k+1}f(n,k-1)=$

$\frac{1}{\binom{n}{k-1}}\frac{1}{n+1} \frac{k}{n-k+1 }=$

$=\frac{(k-1)!k(n-(k-1))!}{n!}\frac{1}{(n+1)(n-(k-1))}=$

$= \frac{1}{\binom{n}{k}}\frac{1}{n+1}$