# On identities from derivatives of high order of the Gamma function and harmonic numbers, likes $\Gamma'(m+1)=m!\left(-\gamma+H_m\right)$

I know that the Gamma function is related with harmonic numbers since in the sum $$1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$$ for $n>1$, one can get particular values of the Gamma function as the denominators of such sum (the numerators thus are the unsigned Stirling numbers of first kind $s(n+1,2):=a(n+1)$, see if you want the Sloane's sequence A000254).

Also I know that particular values of the derivative of the Gamma function are related with harmonic numbers, Wikipedia said that there is a relation between the derivative of the Gamma function and harmonic numbers $$\Gamma'(m+1)=m!\left(-\gamma+H_m\right)$$ (see Properties in the article dedicated to Gamma function).

Question. Is there relation between particular values of a derivative of high order of the Gamma function and harmonic numbers? For example can you do an specilization or the second or third derivative of the Gamma function yielding an identity with harmonic numbers (you can use generalized harmonic numbers). Thanks in advance.

• If there is a well know formula/theorem for the general case of the $nth$ derivative of the Gamma function, you can add the reference. It is appreciated. – user243301 Sep 16 '16 at 10:08
• You can use $\Gamma'(x) = \psi(x)\Gamma(x)$ to express $\Gamma^{(n)}(x)$ with the polygamma functions $\psi^{(n)}(x)$ and $\Gamma(x).$ – gammatester Sep 16 '16 at 10:17
• Then if there is a relation,what's it? Thanks @gammatester . My goal is understand some easy facts about Gamma function and harmonic numbers. – user243301 Sep 16 '16 at 10:20
• I do not know the explicit relation, and the formula from the Wolfram function site functions.wolfram.com/06.05.20.0003.02 seems to be no real help. – gammatester Sep 16 '16 at 10:22
• Look at the proof, and see if it works for higher derivatives : $\qquad$since $\Gamma(s+1) = s \Gamma(s)$, we have $\Gamma'(n+1) = \Gamma(n) + n \Gamma'(n)$. We want to check that $\Gamma'(n+1) = n! (-\gamma+H_{n})$ - Since $\Gamma'(1) = -\gamma$, it is true for $n=0$. - And if it is true for some $n$, then $\Gamma'(n+1) = n! (-\gamma+H_{n}) \implies \Gamma'(n+2) = \Gamma(n+1) + (n+1) \Gamma'(n+1)$ $= n! + (n+1) n! (-\gamma+H_n) = (n+1)! (-\gamma+H_{n+1})$ i.e. it is true for $n+1$. $\implies$ by induction, it is true for every $n \in \mathbb{N}$. – reuns Sep 16 '16 at 10:49

Using $$\lim\limits_{n\to\infty}\frac{\Gamma(n+x)}{n^x\,\Gamma(n)}=1$$, which follows from the recursion $$\Gamma(x+1)=x\,\Gamma(x)$$ and the log-convexity of $$\Gamma(x)$$, we get $$\log(\Gamma(x))=\lim_{n\to\infty}\left[(x-1)\log(n)+\sum_{k=0}^n\log\left(\frac{k+1}{k+x}\right)\right]\tag1$$ Taking the derivative, we get \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\log(\Gamma(x)) &=\lim_{n\to\infty}\left[\log(n)-\sum_{k=0}^n\frac1{k+x}\right]\\ &=-\gamma+\sum_{k=0}^\infty\left(\frac1{k+1}-\frac1{k+x}\right)\\[3pt] &=-\gamma+H_{x-1}\tag2 \end{align} and for $$m\ge2$$, \begin{align} \frac{\mathrm{d}^m}{\mathrm{d}x^m}\log(\Gamma(x)) &=(-1)^m(m-1)!\sum_{k=0}^\infty\frac1{(k+x)^m}\\[6pt] &=(-1)^m(m-1)!\left(\zeta(m)-H^{(m)}_{x-1}\right)\tag3 \end{align} and one can compute the derivatives of $$\Gamma(x)$$ from the derivatives of $$\log(\Gamma(x))$$.