Question Number 147356 by Mrsof last updated on 20/Jul/21
$$\frac{{d}}{{dn}}\left({n}!\right) \\ $$
Answered by puissant last updated on 20/Jul/21
$$\Gamma\left({n}\right)=\left({n}−\mathrm{1}\right)!=\int_{\mathrm{0}} ^{+\infty} {t}^{{n}−\mathrm{1}} {e}^{−{t}} {dt} \\ $$$$\frac{\mathrm{1}}{\Gamma\left({n}\right)}=\:{ne}^{\gamma{n}} \underset{{k}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}+\frac{{n}}{{k}}\right){e}^{−\frac{{n}}{{k}}} \\ $$$${ln}\left(\frac{\mathrm{1}}{\Gamma\left({n}\right)}\right)=−\left({ln}\left(\Gamma\left({x}\right)\right)\right. \\ $$$$={ln}\left({n}\right)+\gamma{n}+\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}{ln}\left(\mathrm{1}+\frac{{n}}{{k}}\right)−\frac{{n}}{{k}}… \\ $$$$\frac{{d}}{{dn}}\left({ln}\left(\frac{\mathrm{1}}{\Gamma\left({n}\right)}\right)\right)=−\frac{\Gamma'\left({n}\right)}{\Gamma\left({n}\right)} \\ $$$$=−\left(\frac{\mathrm{1}}{{n}}+\gamma+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{1}+\frac{{n}}{{k}}}×\frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{{k}}\right) \\ $$$$=−\frac{\mathrm{1}}{{n}}−\gamma+\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{{n}+{k}} \\ $$$$\psi\left({n}\right)=−\gamma+\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{{n}−\mathrm{1}+{k}} \\ $$$$\Rightarrow\:\psi\left({n}+\mathrm{1}\right)=−\gamma+\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{{n}+{k}}.. \\ $$$$\Rightarrow\frac{{d}}{{dn}}\left({n}!\right)=\frac{\Gamma'\left({n}+\mathrm{1}\right)}{\Gamma\left({n}+\mathrm{1}\right)}……… \\ $$
Commented by ArielVyny last updated on 20/Jul/21
$${nice}\:{sir} \\ $$