I have the following formula:

$$h(x)=\sum_{R=1}^m \frac{1}{R+1}-1$$

I have re-expressed this (correctly, I hope!) in terms of the harmonic number $H(m)=\sum_{R=1}^m \frac{1}{R}$:


However, Mathematica insists on simplifying $h(x)$ to


where $\gamma$ is the Euler-Mascheroni constant and $\psi$ is the polygamma function, which various websites tell me is given by $\psi_0(m+2)=\frac{d^1}{d(m+2)^1} \ln \Gamma(m+2)$.

Is there a proof for Mathematica's simplification? And have I summarised it correctly?

I've never worked with gamma functions, so I'd be grateful for help.

  • 2
    $\begingroup$ Not sure which website you are looking in, but the definition is that $$ \psi_0(s) = \frac{d}{ds}\log\Gamma(s). $$ This is related to the harmonic number by $ H(n) = \gamma + \psi_0(n+1)$ for all $n \geq 0$, hence is often used to extend $H$ to non-integral argument. $\endgroup$ – Sangchul Lee Apr 5 '18 at 16:18
  • $\begingroup$ Yeah - wrote $d^0$ instead of $d^1$. Question now edited. So, what is the proof of $H(n)=\gamma+\psi_0(n+1)$ for all $n≥0$? $\endgroup$ – Richard Burke-Ward Apr 5 '18 at 16:24
  • $\begingroup$ Log-differentiating the functional identity $\Gamma(s+1) = s\Gamma(s)$ gives the identity $\psi_0(s+1) = \psi_0(s) + \frac{1}{s}$. Together with the identity $\psi_0(1) = -\gamma$ gives the desired relation. $\endgroup$ – Sangchul Lee Apr 5 '18 at 16:34
  • $\begingroup$ This answer might prove relevant, $\endgroup$ – robjohn Apr 5 '18 at 16:37
  • $\begingroup$ Thank you both. I can't mark this as answered because these are comments, but I do consider this answered. $\endgroup$ – Richard Burke-Ward Apr 5 '18 at 16:41

You can find this in my notes too, in the section about special functions.

The Mittag-Leffler and Weierstrass theorems about the factorization of entire functions equip us with the nice identity $$ \Gamma(z+1) = e^{-\gamma z}\prod_{n\geq 1}\left(1+\frac{z}{n}\right)^{-1} e^{z/n} \tag{1}$$ which holds uniformly over any compact subset of $\mathbb{C}\setminus\{0,-1,-2,-3,\ldots\}$. It can be proved also through Euler's product formula, which on its turn, if restricted to the positive real line, is a consequence of the definition of the $\Gamma$ function provided by the Bohr-Mollerup theorem, i.e. $\Gamma(x+1)$ is the only log-convex function which fulfills the functional equation $f(x+1)=(x+1)\cdot f(x),\;f(0)=1$, i.e. the "most natural" extension of the factorial function to the positive real numbers. The constant $\gamma$ appearing above stands for $$ \lim_{n\to +\infty} H_n-\log(n) = \sum_{n\geq 1}\left(\frac{1}{n}-\log\left(1+\frac{1}{n}\right)\right)\stackrel{\mathcal{L}^{-1}}{=}\int_{0}^{+\infty}\frac{1}{e^x-1}-\frac{1}{x e^x}\,dx\approx\frac{1}{\sqrt{3}}. \tag{2}$$ If we apply $\frac{d}{dz}\log(\cdot)$ to both sides of $(1)$ and define the Digamma function as $\frac{d}{dz}\log\Gamma(z)=\frac{\Gamma'(z)}{\Gamma(z)}$ we have: $$ \psi(z+1)+\gamma = \sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{n+z}\right) \tag{3} $$ and if $z\in\mathbb{N}$ the RHS of $(3)$ is a telescopic series, equal to $H_z$ by the combinatorial definition of harmonic numbers. Clearly $H_z = \gamma+\psi(z+1)$ allows to define harmonic numbers with non-integer parameters, and the functional identities for the $\Gamma$ function $$ \Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin(\pi s)},\qquad \Gamma(x+1)=x\cdot\Gamma(x) $$ imply the following functional identities for the Digamma function, by logarithmic differentiation: $$ \psi(s)-\psi(1-s) = -\pi\cot(\pi s),\qquad \psi(s+1)=\frac{1}{s}+\psi(s). \tag{4}$$ We also have duplication and multiplication formulas, and a little gem from Gauss.
They lead to a number of interesting facts, among them: $$ H_{1/2}=2-2\log 2,\qquad \int_{0}^{+\infty}\frac{dx}{1+x^a}=\frac{\pi}{a\sin\frac{\pi}{a}}\quad\forall a>1. \tag{5}$$ By differentiating $(3)$ and invoking creative telescoping, a short proof of Stirling's approximation follows.

  • $\begingroup$ I would appreciate an explanation for the downvote. $\endgroup$ – Jack D'Aurizio Apr 6 '18 at 11:14
  • $\begingroup$ I believe that everything stated above is true, but perhaps a bit above the level of the question. In any case, how does this answer the question, which asks about Mathematica's simplification of the expression given? $\endgroup$ – robjohn Apr 8 '18 at 13:57
  • $\begingroup$ @robjohn: your $(15)$ is my $(3)$. $\endgroup$ – Jack D'Aurizio Apr 8 '18 at 16:34
  • $\begingroup$ Your $(3)$ is closer to my $(12)$, and my $(12)$ addresses the question in the title, but not the question in the body. BTW, I was not the downvoter. $\endgroup$ – robjohn Apr 8 '18 at 22:07

The Digamma Function

Differentiating the logarithm of $\Gamma(x+1)=x\,\Gamma(x)$ gives $$ \frac{\Gamma'(x+1)}{\Gamma(x+1)}=\frac1x+\frac{\Gamma'(x)}{\Gamma(x)}\tag1 $$ that is, the digamma function satisfies $$ \psi(x+1)=\frac1x+\psi(x)\tag2 $$ Since $\log(\Gamma(x+1))-\log(\Gamma(x))=\log(x)$, the Mean Value Theorem says that there is a $\xi\in(x,x+1)$ so that $$ \frac{\Gamma'(\xi)}{\Gamma(\xi)}=\log(x)\tag3 $$ Because $\Gamma$ is log-convex, we get that $$ \log(x-1)\lt\psi(x)\lt\log(x)\tag4 $$ Therefore, $$ \lim_{x\to\infty}(\psi(x)-\log(x))=0\tag5 $$

The Extended Harmonic Numbers

If we define $$ H(x)=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+x}\right)\tag6 $$ then we get by using Telescoping Series $$ \begin{align} H(x+1)-H(x) &=\sum_{k=1}^\infty\left(\frac1{k+x}-\frac1{k+x+1}\right)\\ &=\frac1{x+1}\tag7 \end{align} $$ and for $n\in\mathbb{N}$, $H(n)=H_n$, the $n^\text{th}$ Harmonic Number.

As is shown in this answer, $$ \lim_{n\to\infty}(H_n-\log(n))=\gamma\doteq0.57721566490153286060651209\tag8 $$ If $|x-n|\lt1$, then $|H_n-H(x)|\le\frac1{n-1}$ and $|\log(n)-\log(x)|\le\frac1{n-1}$. Therefore, we get that $$ \lim_{x\to\infty}(H(x)-\log(x))=\gamma\tag9 $$

Relation Between $\boldsymbol{H(x)}$ and $\boldsymbol{\psi(x+1)}$

Equations $(2)$ and $(7)$ say that for $f(x)=H(x)-\psi(x+1)$, we have $$ f(x+1)=f(x)\tag{10} $$ Equations $(5)$ and $(9)$ say that $$ \lim_{x\to\infty}f(x)=\gamma\tag{11} $$ Therefore, we get that $f(x)=\gamma$; that is, $$ H(x)=\psi(x+1)+\gamma\tag{12} $$

To The Question

In your question you seem to be equating $x$ and $m$. I will use only $m$ since it is more commonly used for an integer.

Your equation $$ h(m)=H_m+\frac1{m+1}-2\tag{13} $$ is correct. $(13)$ can also be written as $$ h(m)=H_{m+1}-2\tag{14} $$ As shown in $(12)$, $H_{m+1}=\psi(m+2)+\gamma$, so $(14)$ becomes $$ h(m)=\psi(m+2)+\gamma-2\tag{15} $$ It would seem that the formula that Mathematica gives would be for $$ \gamma+\psi(m+2)-H_m-2=h(m)-H_m\tag{16} $$


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