Prove that $\sum\limits_{k=0}^n (-1)^k\:{\binom n k}^{-1}=\frac{n+1}{n+2} (1+(-1)^n)$ $$\sum _{k=0}^n (-1)^k \frac{1}{\binom n k}=\frac{n+1}{n+2} (1+(-1)^n)$$
$$A(n,k)=(-1)^k {\binom n k}^{-1}=(-1)^k \frac{(n-k)!k!}{n!}$$
$$A(n+1,k+1)-A(n+1,k)=-\frac{n+2}{n+1} A (n,k)$$
$$\sum_{k=1}^n A(n,k)=-\frac{n+1}{n+2} (A(n+1,n+1)-A(n+1,0)) = (1+(-1)^n) \: \frac{n+1}{n+2}$$
Is my approach correct?
 A: Another approach
$$
\begin{align}
\sum_{k=0}^n\frac{(-1)^k}{\binom{n}{k}}
&=(n+1)\sum_{k=0}^n(-1)^k\frac{\Gamma(k+1)\Gamma(n-k+1)}{\Gamma(n+2)}\tag{1}\\
&=(n+1)\sum_{k=0}^n(-1)^k\int_0^1t^k(1-t)^{n-k}\,\mathrm{d}t\tag{2}\\
&=(n+1)\int_0^1\frac{1-(-1)^{n+1}\frac{t^{n+1}}{(1-t)^{n+1}}}{1+\frac{t}{1-t}}(1-t)^n\,\mathrm{d}t\tag{3}\\
&=(n+1)\int_0^1\left((1-t)^{n+1}-(-1)^{n+1}t^{n+1}\right)\,\mathrm{d}t\tag{4}\\[4pt]
&=\frac{n+1}{n+2}\left(1-(-1)^{n+1}\right)\tag{5}
\end{align}
$$
Explanation:
$(1)$: write the binomial coefficients as ratios of Gamma Functions
$(2)$: Use the Beta Function integral
$(3)$: apply the Formula for the Sum of a Geometric Series
$(4)$: simplify
$(5)$: integrate
A: Your approach is nice and correct. It's a good idea to simplify the sum with the technique of telescoping.

Two aspects:

*

*Since your proof is valid for all non-negative integers $n$ you should clearly state the range of validity.


*Since the approach of telescoping is valid and sufficient to prove the claim no proof by induction is necessary.

Typo: The sum in the last line should start with $k=0$.
