probability - (Combinatorial?) Proof of the identity $sum_{k=1}^n frac {(-1)^k}{k,(k+1)}binom nk = frac 12 + frac 13 + dots + frac 1{n+1}$?

Recently I've come across an interesting identity:
$$\frac 1{1\cdot 2}\binom n1 - \frac 1{2\cdot 3}\binom n2 + \frac 1{3\cdot 4}\binom n3 - \dots + \frac {(-1)^n}{n\cdot (n+1)}\binom nn = \frac 12 + \frac 13 + \dots + \frac 1{n+1}.$$

Does anyone have an idea how to prove this?

PS. It'd be of special interest if someone could provide a combinatorial approach to this problem, although I'm not sure if that'd make sense since this is an identity for non-integer.

Edit: Perhaps my question was worded in a way that seems trivial since there's a vote to close this question. I should have mentioned that I can prove it by expanding $\frac 1{k(k-1)}$ and do some further calculation. What I want is a clever way to interpret it, e.g. using combinatorics or some kind of generating function or integration.

Answer

For the solution see original post below

EDIT

Here are direct proofs of (1), (2), and (3), each in one sequence

ad (1)

$$H_{n} = \sum_{k=1}^n \frac{1}{k} = \sum_{k=1}^n \int_0^1 y^{k-1}= \int_0^1\,dy \sum_{k=1}^n y^{k-1}= \int_0^1\,dy \frac{1-y^n}{1-y} \\({y\to 1-x})=\int_0^1\,dx \frac{1}{x} \left(1-(1-x)^n\right)= \int_0^1\,dx \sum_{k=1}^n (-1)^{k+1}\binom{n}{k} x^{k-1} \\= \sum_{k=1}^n (-1)^{k+1}\binom{n}{k}\int_0^1\,dx\; x^{k-1} = \sum_{k=1}^n (-1)^{k+1}\binom{n}{k}\frac{1}{k}$$

QED.

ad (2)

$$\sum _{k=1}^n \frac{(-1)^{k+1} }{k+1}\binom{n}{k}=\sum _{k=1}^n (-1)^{k+1}\binom{n}{k} \int_0^1 x^k \,dx =\int_0^1 \,dx \sum _{k=1}^n (-1)^{k+1}\binom{n}{k} x^k \\= \int_0^1 \, dx \left(1-(1-x)^n\right)= 1-\frac{1}{n+1} = \frac{n}{n+1}$$

QED.

ad (3)

$$\sum _{k=1}^n (-1)^{k+1} \binom{n}{k} \frac{1}{a+k}=\sum_{k=1}^n (-1)^{k+1} \binom{n}{k} \int_0^1 \,dx x^{a+k-1}\\=\int_0^1\,dx x^{a-1} \sum _{k=1}^n (-1)^{k+1} \binom{n}{k} x^{k}=\int_0^1\,dx x^{a-1} \left(1-(1-x)^n\right)\\= \frac{1}{a} - \int_0^1\,dx x^{a-1} (1-x)^n= \frac{1}{a} - \text{Beta}(a,n+1)= \frac{1}{a} - \frac{\Gamma(a) \Gamma(n+1)}{\Gamma(a+n+1)}$$

QED.

Remark

Combining (3) and (1) we see that

$$H_{n} = \lim_{a\to0}\left( \frac{1}{a} - \frac{\Gamma(a) \Gamma(n+1)}{\Gamma(a+n+1)}\right)\tag{4a}$$

To make this explicit we write the r.h.s as

$$\frac{1}{a} \left( 1- \frac{\Gamma(a+1) \Gamma(n+1)}{\Gamma(a+n+1)}\right)=\frac{\Gamma(a+n+1)-\Gamma(a+1) \Gamma(n+1)}{a\Gamma(a+n+1)}$$

and then apply l'Hospital to arrive at

$$H_{n} =\frac{ \frac{\partial (\Gamma (a+n+1)-n! \Gamma (a+1))}{\partial a}\text{/.}\, a\to 0} {\frac{\partial (a \Gamma (a+n+1))}{\partial a}\text{/.}\, a\to 0} = \frac{ \Gamma' (n+1)-n! \Gamma' (1)} {\Gamma(n+1)}\\= \frac{ \Gamma' (n+1)}{\Gamma(n+1)}+ \gamma = \frac{\partial \log (\Gamma (n+1))}{\partial n}+\gamma= \psi ^{(0)}(n+1)+\gamma\tag{4b}$$

Here $\gamma= -\Gamma'(1)$ is the Euler-Mascheroni constant and $\psi$ is the polygamma function.

Original post

This is not the requested combinatorical interpretation but there are some interesting identities popping up.

Subtracting these two interesting relations

$$\sum _{k=1}^n \frac{(-1)^{k+1}}{k} \binom{n}{k} = H_{n}\tag{1}$$

and

$$\sum _{k=1}^n \frac{(-1)^{k+1}}{1+k} \binom{n}{k} = \frac{n}{1+n}\tag{2}$$

and observing

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

we find that the left hand side is that of the formula we have to prove while the right hand side gives

$$H_{n}-\frac{n}{1+n} = H_{n+1}-1$$

which proves the OP.

Here we have used the basic relation

$$H_{n}+\frac{1}{1+n} =H_{n+1}$$

Remarks

1) I found it surprising that the small difference in the l.h.s. of (1) and (2) leads to a apreciably different results.

2) formulas (1) and (2) can be generalized to

$$\sum _{k=1}^n \frac{(-1)^{k+1}}{a+k} \binom{n}{k} = \frac{1}{a}-\frac{\Gamma (a) \Gamma (n+1)}{\Gamma (a+n+1)}\tag{3}$$