Derivative of binomial coefficients I obtained the following formula in Mathematica:
$$\frac{d}{dn}\ln\binom{n}{k} = H_{n} - H_{n-k}$$
where $H_n$ are the harmonic numbers ($H_n = \sum_{i=1}^n 1/i$). But I have no idea how to prove it. Can someone help me? Or at least provide a reference to a textbook/paper?
 A: \begin{eqnarray*}
\binom{n}{k} &=& \frac{n(n-1) \cdots(n-k+1)}{k!} \\
\ln \binom{n}{k} &=&\ln n + \ln(n-1) + \cdots +\ln(n-k+1) -\ln(k!) \\
\frac{d}{dn} \ln \binom{n}{k} &=& \frac{1}{n} + \frac{1}{n-1} +\cdots + \frac{1}{n-k+1} =\color{red}{H_n-H_{n-k}}.
\end{eqnarray*}
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\ds{\totald{}{n}\ln\pars{n \choose k}}} =
\totald{}{n}\bracks{\ln\pars{\Gamma\pars{n + 1}} -
\ln\pars{\Gamma\pars{n - k + 1}}}\quad\pars{~\Gamma:\ Gamma Function~}
\\[5mm] = &\
\Psi\pars{n + 1} - \Psi\pars{n + 1 - k}\qquad\pars{~\Psi:\ Digamma Function~}
\\[5mm] = &\
\underbrace{\bracks{\Psi\pars{n + 1} + \gamma}}
_{\ds{H_{n}}}\ -\
\underbrace{\bracks{\Psi\pars{n - k + 1} + \gamma}}_{\ds{H_{n - k}}}
\qquad \pars{~\gamma:\ Euler\!-\!Mascheroni\ Constant~}
\\[5mm] = &\ \bbx{H_{n} - H_{n - k}}
\end{align}
A: $\frac{d}{dn}\ln{n\choose k}=\frac{d}{dn}[\ln(n!)-\ln((n-k)!)-\ln(k!)]=\frac{d}{dn}\ln(n!)-\frac{d}{dn}\ln((n-k)!)$
Can you take it from here?
(Hint: $\ln(\Pi a_i)=\sum_i \ln(a_i)$)
A: $$k!\binom{n}k=n(n-1)(n-2)\cdots(n-k+1).$$
Taking the logarithmic derivative with respect to $n$ gives
$$\frac1n+\frac1{n-1}+\frac1{n-2}+\cdots+\frac1{n-k+1}$$
(as logarithmic derivative converts products into sums).
