<s>[INST] Coefficient extraction I want to show:
\begin{equation*}
  [z^n]\frac{1}{(1-z)^{\alpha + 1}} \log \frac{1}{1-z} = \binom{n + \alpha }{n} (H_{n+\alpha} - H_{\alpha}).
\end{equation*}
where $[z^n]$ means the $n$-th coefficient of the power series and
\begin{equation}
H_{n+\alpha} - H_{\alpha} = \sum^{n}_{k=1}{\frac{1}{\alpha + k}}
\end{equation}
So far I got
\begin{equation*}
    \frac{1}{(1 - z)^{\alpha + 1}} = (1-z)^{-(\alpha + 1)} = \sum_{n \geq 0}{\binom{-\alpha - 1}{k}(-1)^k z^k} =  \sum_{n \geq 0}{\binom{\alpha + n }{n} z^n}
  \end{equation*}
where I used $\binom{-\alpha}{k} = (-1)^k \binom{\alpha + k - 1}{k}$and
\begin{equation*}
    \log \frac{1}{1 - z} = - \log 1- z = \sum_{n \geq 1}\frac{z^n}{n} = z\sum_{n \geq 0}\frac{z^{n}}{n+1}
  \end{equation*}
therefore
\begin{align*}
    \frac{1}{(1-z)^{\alpha + 1}} \log \frac{1}{1-z} &= z\left( \sum_{n \geq 0}{\binom{\alpha + n }{n} z^n} \right) \left( \sum_{n \geq 0}\frac{z^{n}}{n+1} \right) \\
    &= z \sum_{n \geq 0}\sum^{n}_{k=0}\binom{\alpha + k}{k} \frac{1}{n - k + 1} z^{n} \\
    &= \sum_{n \geq 0}\sum^{n}_{k=0}\binom{\alpha + k}{k} \frac{1}{n - k + 1} z^{n + 1}.
  \end{align*}
Now the $n$-th coefficient is
\begin{align*}
    \sum^{n}_{k=0}\binom{\alpha + k}{k} \frac{1}{n - k + 1} &= \binom{\alpha + n}{n} \sum^n_{k = 1}{\frac{n!}{(n-k)! k!} \frac{1}{\binom{\alpha + n}{n - k}}    \frac{1}{n - k + 1}} +  \frac{1}{n+1}\\
    &=  \binom{\alpha + n}{n} \sum^n_{k = 1}{\binom{n}{k} \frac{1}{\binom{\alpha + n}{n - k}}    \frac{1}{n - k + 1}}  +   \frac{1}{n+1}\\
    &= \binom{\alpha + n}{n} \sum^n_{k = 1}{\binom{n}{k - 1} \frac{1}{\binom{\alpha + n}{n - k - 1}} \frac{k-1}{n - k -1}  \frac{1}{\alpha + k + 1} } + \frac{1}{n+1}\\
    &= \binom{\alpha + n}{n} \sum^{n+1}_{k = 2}{\binom{n}{k - 2} \frac{1}{\binom{\alpha + n}{n - k}} \frac{k- 2 }{n - k}  \frac{1}{\alpha + k} } + \frac{1}{n+1}
  \end{align*}
but from now on I can't see how to proceed.
I also know that
\begin{equation}
[z^n] \frac{1}{1-z} \log \frac{1}{1-z} = H_n
\end{equation}
somehow I also tried to use some sort of transformation law for
coefficient extraction, but I am not aware of any kind of transformation law for coefficient extraction. Using $1-u = (1-z)^{\alpha + 1}$ or $z = 1 - (1-u)^{1/(\alpha + 1)}$gives
\begin{equation}
\frac{1}{\alpha + 1}\frac{1}{1-u} \log \frac{1}{1 - u}
\end{equation}
and for this I get the coefficient $\frac{1}{\alpha + 1} H_n$.
Another approach I did was the following:
Since
\begin{equation*}
    (1-z)^{-m} = \exp(-m \ln(1-z)) 
  \end{equation*}
I get
\begin{equation*}
    \frac{\partial }{\partial m} \exp(-m \ln(1-z))  = -\ln(1-z)  \exp(- m \ln(1-z)) = - \ln(1-z) \frac{1}{(1-z)^m} = \frac{1}{(1-z)^m} \ln \frac{1}{1-z}
  \end{equation*}
therefore
\begin{align*}
    \frac{\partial }{\partial m} (1-z)^{-m} &= \sum_{n \geq 0}{\frac{\partial }{\partial m} \binom{m + n - 1}{n} z^n} \\
    &= \sum_{n \geq 0}{\frac{\partial }{\partial m} \frac{\Gamma(m + n)}{n!\Gamma(m)} z^n} 
\end{align*}
But I don't know any identities for the derivative of the gamma function.
 [/INST] Here is another variation. We show the identity
\begin{align*}
\color{blue}{[z^n]\frac{1}{(1-z)^{\alpha+1}}\log\frac{1}{1-z}=\binom{n+\alpha}{n}\sum_{k=1}^n\frac{1}{\alpha+k}}\tag{1}
\end{align*}
is valid for integral $n>0$ and $\alpha \in\mathbb{C}\setminus\{-1,\ldots,-n\}$.

We start with the LHS of (1) and obtain
\begin{align*}
\color{blue}{[z^n]}\color{blue}{\frac{1}{(1-z)^{\alpha+1}}\log\frac{1}{1-z}}
&=[z^n]\frac{1}{(1-z)^{\alpha+1}}\sum_{k=1}^\infty\frac{z^k}{k}\tag{2.1}\\
&=\sum_{k=1}^n\frac{1}{k}[z^{n-k}]\frac{1}{(1-z)^{\alpha+1}}\tag{2.2}\\
&=\sum_{k=1}^n\frac{1}{k}\binom{-\alpha-1}{n-k}(-1)^{n-k}\tag{2.3}\\
&\,\,\color{blue}{=\sum_{k=1}^n\frac{1}{k}\binom{n+\alpha-k}{n-k}}\tag{2.4}
\end{align*}

Comment:

*

*In (2.1) we use the logarithmic series expansion.


*In (2.2) we apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$ and restrict the upper limit of the sum to $n$, since other terms do not contribute.


*In (2.3) we select the coefficient of $[z^{n-k}]$ of the binomial series expansion.


*In (2.4) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

It remains to show the binomial identity
\begin{align*}
\color{blue}{\sum_{k=1}^n\frac{1}{k}\binom{n+\alpha-k}{n-k}=\binom{n+\alpha}{n}\sum_{k=1}^n\frac{1}{\alpha+k}}\tag{3}
\end{align*}

A nice aspect is, that when considering the expressions in (3) as polynomial in $\alpha$ we can show, that they both are the derivative $\frac{\partial}{\partial\alpha}\binom{n+\alpha}{n}$.

We start with the RHS of (3) which is the easy part. We obtain
\begin{align*}
\color{blue}{\frac{\partial}{\partial \alpha}}\color{blue}{\binom{n+\alpha}{n}}
&=\frac{1}{n!}\frac{\partial}{\partial \alpha}\prod_{j=0}^{n-1}(n+\alpha-j)\tag{4.1}\\
&=\frac{1}{n!}\frac{\partial}{\partial \alpha}\prod_{j=0}^{n-1}(\alpha+j+1)\tag{4.2}\\
&=\frac{1}{n!}\frac{\partial}{\partial \alpha}\prod_{j=1}^{n}(\alpha+j)\tag{4.3}\\
&=\frac{1}{n!}\sum_{k=1}^n\left(\frac{\partial}{\partial \alpha}(\alpha+k)\right)\prod_{{j=1}\atop {j\ne k}}^n(\alpha+j)\tag{4.4}\\
&=\frac{1}{n!}\sum_{k=1}^n\frac{1}{\alpha+k}\prod_{j=1}^n(\alpha+j)\tag{4.5}\\
&\,\,\color{blue}{=\binom{n+\alpha}{n}\sum_{k=1}^n\frac{1}{\alpha+k}}
\end{align*}
and we get the RHS of (3).

Comment:

*

*In (4.1) we use the definition of binomial coefficients in the form $\binom{p}{q}=\frac{1}{q!}\prod_{j=0}^{q-1}(p-j)$.


*In (4.2) we change the order of multiplication $j\to n-1-j$.


*In (4.3) we shift the index and start with $j=1$.


*In (4.4) we use the product rule for derivation of products with $n$ terms.


*In (4.5) we do the derivation and expand numerator and denominator with $\alpha+k$.

Now the somewhat more challenging part. We do the derivation by using first principles. We obtain
\begin{align*}
\color{blue}{\frac{\partial}{\partial \alpha}}&\color{blue}{\binom{n+\alpha}{n}}
=(-1)^n\frac{\partial}{\partial \alpha}\binom{-\alpha-1}{n}\tag{5.1}\\
&=(-1)^n\lim_{q\to 0}\frac{1}{q}\left(\binom{-\alpha-1-q}{n}-\binom{-\alpha-1}{n}\right)\tag{5.2}\\
&=(-1)^n\lim_{q\to 0}\frac{1}{q}\left(\sum_{k=0}^n\binom{-q}{k}\binom{-\alpha-1}{n-k}-\binom{-\alpha-1}{n}\right)\tag{5.3}\\
&=(-1)^n\lim_{q\to 0}\frac{1}{q}\sum_{k=1}^n\binom{-q}{k}\binom{-\alpha-1}{n-k}\\
&=(-1)^n\lim_{q\to 0}\frac{1}{q}\sum_{k=1}^n\frac{-q}{k}\binom{-q-1}{k-1}\binom{-\alpha-1}{n-k}\tag{5.4}\\
&=(-1)^{n+1}\sum_{k=1}^n\frac{1}{k}\binom{-1}{k-1}\binom{-\alpha-1}{n-k}\\
&=\sum_{k=1}^n\frac{(-1)^{n-k}}{k}\binom{-\alpha-1}{n-k}\tag{5.5}\\
&\,\,\color{blue}{=\sum_{k=1}^n\frac{1}{k}\binom{n+\alpha-k}{n-k}}\tag{5.6}
\end{align*}
which is the LHS of (3) and the claim (3) follows.

Comment:

*

*In (5.1) we use again the binomial identity as in (2.4).


*In (5.2) we do the derivation according to first principles.


*In (5.3) we apply the Chu-Vandermonde identity.


*In (5.4) we use the binomial identity $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$.


*In (5.5) we use the binomial identity $\binom{-1}{q}=\frac{1}{q!}(-1)(-2)\cdots(-q)=(-1)^q$.


*In (5.6) we use again the binomial identity as in (2.4).
</s>