$$\sum_{k=0}^{n} \binom{n}{k}^{-1} = \frac{n+1}{2^{(n+1)}} \sum_{k=1}^{n+1} \frac{2^k}{k}$$ I think this is a fascinatig equality. I have tried to use this equation first: $${n \choose k}^{-1}=(n+1)\int_{0}^{1} x^k (1-x)^{n-k}~ dx$$
$(n+1)$s are cancelled at both sides: $$\sum_{k=0}^{n}{n \choose k}^{-1}= \sum_{k=0}^{n} \int_{0}^{1} x^k (1-x)^{n-k}~ dx$$ how to apply change of variables so that we can compute $$\sum_{k=0}^{n} x^k (1-x)^{n-k}$$ $x = \sin^2{t} , dx = \sin{2t}dt$ didnt work for me. I have no clue about how to deal with the right side of the equality
