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Question Number 147356 by Mrsof last updated on 20/Jul/21

$$\frac{d}{dn}(n!)$$

Answered by puissant last updated on 20/Jul/21

$$\mathrm{\Gamma }\left(n\right)=(n-1)!={\int }_0^{+\mathrm{\infty }}{t}^{n-1}{e}^{-t}dt$$$$\frac{1}{\mathrm{\Gamma }\left(n\right)}={ne}^{\gamma n}\underset{k=1}{\prod ^{\mathrm{\infty }}}(1+\frac{n}{k})e^{-\frac{n}{k}}$$$$ln\left(\frac{1}{\mathrm{\Gamma }\left(n\right)}\right)=-(ln\left(\mathrm{\Gamma }\left(x\right)\right)$$$$=ln\left(n\right)+\gamma n+\underset{k=1}{\sum ^{\mathrm{\infty }}}ln(1+\frac{n}{k})-\frac{n}{k}...$$$$\frac{d}{dn}\left(ln\left(\frac{1}{\mathrm{\Gamma }\left(n\right)}\right)\right)=-\frac{{\mathrm{\Gamma }}^{\prime }\left(n\right)}{\mathrm{\Gamma }\left(n\right)}$$$$=-(\frac{1}{n}+\gamma +\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{1+\frac{n}{k}}\times \frac{1}{k}-\frac{1}{k})$$$$=-\frac{1}{n}-\gamma +\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k}-\frac{1}{n+k}$$$$\psi \left(n\right)=-\gamma +\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k}-\frac{1}{n-1+k}$$$$\Rightarrow \psi (n+1)=-\gamma +\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k}-\frac{1}{n+k}..$$$$\Rightarrow \frac{d}{dn}(n!)=\frac{{\mathrm{\Gamma }}^{\prime }(n+1)}{\mathrm{\Gamma }(n+1)}.........$$

Commented by ArielVyny last updated on 20/Jul/21

$$nicesir$$

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