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For any $\alpha \in \mathbb{C}$, by studying the $n$-th coefficient (with respect to the variable $z$) of the power series of the function

$$ \frac{1}{(1-z)^{\alpha+1}}\,\log\left(\frac{1}{1-z}\right) $$

in two different ways, we can obtain the identity

$$ \forall n \geqslant 0, \, \forall\alpha \in \mathbb{C}, \;\; \sum_{k=1}^{n} \frac{1}{k} \, \binom{n-k+\alpha}{n-k} \; = \; \binom{n+\alpha}{n}\,\sum_{k=1}^{n} \frac{1}{k+\alpha}.$$

I would like to know if there exists a "direct" proof of this equality, that is, for example, by replacing the binomial coefficients of one of the two members with another expression or something else. Thank you in advance !

Kermatoni
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    The right hand side of your identity is $\dbinom{n+\alpha}{n}$ differentiated by $\alpha$. So you'll probably want a formula for the derivative of a binomial coefficient, such as Theorem 1 in https://math.stackexchange.com/questions/1941994/derivative-of-the-binomial-binom-x-n-with-respect-to-x/2974977#2974977 . – darij grinberg Nov 29 '19 at 18:52
  • The left hand side of your identity cannot be correct as stated, since it is $0$ when $\alpha$ is not a nonnegative integer. Maybe you want $\dbinom{n-k+\alpha}{n-k}$ instead? – darij grinberg Nov 29 '19 at 18:53
  • @darijgrinberg you're right for the left hand side, I have changed, sorry ! For the right hand side, your link for the derivative by $\alpha$ is perfect to prove this identity, thanks, but it looks like the proof about the study of the function in my question. I'm searching an other kind of proof. – Kermatoni Nov 29 '19 at 19:04

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