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

Question

I would like if possible to have a proof of yet another theorem of Binomial Coefficients. This time it is $$\sum_{k=0}^n{(-1)^k\over k+1}\binom{n}{k}={1\over n+1}$$

This arises in a proof of the theorem to the effect that $$\operatorname{\Gamma}(n,\mu)$$ is equal to the hypervolume cut out of the positive unit hyperbox with one corner at the origin and each edge parallel to an axis by the hypersurface defined by $$\prod_{k=1}^n x_k=e^{-\mu} .$$

Again, because I did find a proof of the this theorem other than by the route inwhich this theorem of binomial coefficients arose, I now do effectively have a proof of this theorem of binomial coefficients ... but again, taken as a proof per se it's absurdly complex; and I think there must be an elementary one: and besides, as I have said elsewhere, I like to have an entire map of a mathematical 'landscape' rather than just knowledge of a particular 'route' connecting this place & that.

I was shown two excellent proofs for the theorem of binomial coefficients & Pochhammer numbers that I requested a proof of; bur I don't think these are applicable to this one - not unadapted in someway anyhow - and possibly a proof by a completely different route would serve better in this case.

Answer 1

If you mean $$ \sum_{k=0}^n\frac{(-1)^k}{k+1}\binom{n}{k}=\frac1{n+1} $$ instead of $$ \sum_{k=0}^n\frac{(-1)^n}{k+1}\binom{n}{k}=\frac1{n+1}, $$ (which doesn't make any sense when $n$ is odd, the LHS is negative due to the $(-1)^n$ but the RHS is positive) then one obvious way is just integrating $$ (1-x)^n=\sum_{k=0}^n(-1)^k\binom{n}{k}x^k $$ term-by-term between $x=0$ and $x=1$.

Answer 2

$$ \frac{1}{k+1}\binom{n}{k} = \frac{1}{n+1}\binom{n+1}{k+1} $$ so $$ \sum_{k=0}^n{(-1)^k\over k+1}\binom{n}{k}= \sum_{k=0}^n{(-1)^k\over n+1}\binom{n+1}{k+1}= \frac{1}{n+1}\sum_{k=0}^n{(-1)^k}\binom{n+1}{k+1}= \frac{1}{n+1}[(1-1)^{n+1} + 1] = \frac{1}{n+1}. $$