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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?

a06e
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    Did you try writing the binomial coefficient and expanding using the logarithm? – Clayton Mar 02 '18 at 21:01
  • @Clayton Yes, right after I hit post. But I get discrete sums, where the $n$ appears as a limit in the sum. I don't think that's the way to go if I want to differentiate with respect to $n$ – a06e Mar 02 '18 at 21:09
  • Silly question: When you say you obtained the identity in Mathematica(I assume Worlfram Mathematica) you mean you verified it in Mathematica? – BigM Mar 02 '18 at 21:11
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    @BigM I mean that if you differentiate this in Mathematica and FullSimplify it, this is what you get. – a06e Mar 02 '18 at 21:11
  • Nice formula. Possibly related https://math.stackexchange.com/questions/2668419/binomial-harmonic-numbers/2673691#2673691 – Dr. Wolfgang Hintze Mar 03 '18 at 07:49
  • Differentiating with respect to the second parameter gives similarly $\frac{d }{dk} \log \binom{n}{k}=H_{n-k}-H_k$ – Dr. Wolfgang Hintze Mar 03 '18 at 08:43

4 Answers4

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\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*}

Donald Splutterwit
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  • Is the sum formula for the logarithm of the binomial coefficient valid when $n$ is not integer? – a06e Mar 02 '18 at 21:24
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    The first equation above can be used to define a binomial coefficient when $n$ is not an integer ... ( maybe a real or complex number). – Donald Splutterwit Mar 02 '18 at 21:27
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

Felix Marin
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$\frac{d}{dn}\ln{n\choose k}=\frac{d}{dn}[\ln(n!)-\ln((n-k)!)-\ln(k!)]=\frac{d}{dn}\ln(n!)-\frac{d}{dn}\ln((n-k)!)$ Can you take it from here?

(Hint: $\ln(\Pi a_i)=\sum_i \ln(a_i)$)

  • But how do you differentiate $\sum_{i=1}^n \ln(i)$ with respect to $n$? – a06e Mar 02 '18 at 21:14
  • @becko Just write is as $\sum_i \ln(n-i)$ –  Mar 02 '18 at 21:15
  • I see. That's clever. But it's a trick and I am not 100% sure about it. You still have an $n$ in the upper limit in the sum. You just ignore it? – a06e Mar 02 '18 at 21:16
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$$k!\binom{n}k=n(n-1)(n-2)\cdots(n-k+1).$$ 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).

Angina Seng
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  • Is the sum formula for the logarithm of the binomial coefficient valid when n is not integer? – a06e Mar 02 '18 at 21:24
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    @becko It's horrid to use (as you did) $n$ for a real variable, but the formula in my first line is often used to define $\binom xk$ for real $x$, and all the above is true in this generality. – Angina Seng Mar 02 '18 at 21:27