Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 124548 by mnjuly1970 last updated on 04/Dec/20

                ...nice  calculus...         evaluate :::  lim_(x→0) {(1/x)[((ln(Γ(1+x))/x)−ψ(x+1)]}=?

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...{nice}\:\:{calculus}... \\ $$$$\:\:\:\:\:\:\:{evaluate}\:::: \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \left\{\frac{\mathrm{1}}{{x}}\left[\frac{{ln}\left(\Gamma\left(\mathrm{1}+{x}\right)\right.}{{x}}−\psi\left({x}+\mathrm{1}\right)\right]\right\}=? \\ $$$$ \\ $$

Answered by mindispower last updated on 04/Dec/20

ln(Γ(1+x))=f(x)  f(x)=f(0)+f′(0)x+((f′′(0)x^2 )/2)+o(x^2 )  f(0)=0,f′(0)=Ψ(1),f′′(0)=Ψ^1 (1)  Ψ(x+1)=Ψ(1)+Ψ^1 (1)x+o(x)  ((ln(Γ(1+x)))/x)−Ψ(x+1)=((xΨ(1)+(x^2 /2)Ψ^1 (1)+o(x^2 ))/x)−Ψ(1)−Ψ^1 (1)x+o(x)  =−((Ψ^1 (1))/2)x+o(x)  lim_(x→0) {(1/x)[((ln(Γ(1+x)))/x)−Ψ(1+x)]}=lim_(x→0) (1/x).((−Ψ(1))/2)x+o(x)  =lim_(x→0) −((Ψ^1 (1))/2)+o(1)=−((Ψ^1 (1))/2)  Ψ^1 (z)=Σ_(j≥0) (1/((j+z)^2 ))⇒Ψ^1 (1)=Σ_(j≥0) (1/((1+j)^2 ))=(π^2 /6)  we get −(π^2 /(12))

$${ln}\left(\Gamma\left(\mathrm{1}+{x}\right)\right)={f}\left({x}\right) \\ $$$${f}\left({x}\right)={f}\left(\mathrm{0}\right)+{f}'\left(\mathrm{0}\right){x}+\frac{{f}''\left(\mathrm{0}\right){x}^{\mathrm{2}} }{\mathrm{2}}+{o}\left({x}^{\mathrm{2}} \right) \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0},{f}'\left(\mathrm{0}\right)=\Psi\left(\mathrm{1}\right),{f}''\left(\mathrm{0}\right)=\Psi^{\mathrm{1}} \left(\mathrm{1}\right) \\ $$$$\Psi\left({x}+\mathrm{1}\right)=\Psi\left(\mathrm{1}\right)+\Psi^{\mathrm{1}} \left(\mathrm{1}\right){x}+{o}\left({x}\right) \\ $$$$\frac{{ln}\left(\Gamma\left(\mathrm{1}+{x}\right)\right)}{{x}}−\Psi\left({x}+\mathrm{1}\right)=\frac{{x}\Psi\left(\mathrm{1}\right)+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\Psi^{\mathrm{1}} \left(\mathrm{1}\right)+{o}\left({x}^{\mathrm{2}} \right)}{{x}}−\Psi\left(\mathrm{1}\right)−\Psi^{\mathrm{1}} \left(\mathrm{1}\right){x}+{o}\left({x}\right) \\ $$$$=−\frac{\Psi^{\mathrm{1}} \left(\mathrm{1}\right)}{\mathrm{2}}{x}+{o}\left({x}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\frac{\mathrm{1}}{{x}}\left[\frac{{ln}\left(\Gamma\left(\mathrm{1}+{x}\right)\right)}{{x}}−\Psi\left(\mathrm{1}+{x}\right)\right]\right\}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{{x}}.\frac{−\Psi\left(\mathrm{1}\right)}{\mathrm{2}}{x}+{o}\left({x}\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}−\frac{\Psi^{\mathrm{1}} \left(\mathrm{1}\right)}{\mathrm{2}}+{o}\left(\mathrm{1}\right)=−\frac{\Psi^{\mathrm{1}} \left(\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\Psi^{\mathrm{1}} \left({z}\right)=\underset{{j}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left({j}+{z}\right)^{\mathrm{2}} }\Rightarrow\Psi^{\mathrm{1}} \left(\mathrm{1}\right)=\underset{{j}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{1}+{j}\right)^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$${we}\:{get}\:−\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$

Commented by mnjuly1970 last updated on 04/Dec/20

bravo mr mindspower   maclauren expansion of  Γ(x+1)  grateful...

$${bravo}\:{mr}\:{mindspower} \\ $$$$\:{maclauren}\:{expansion}\:{of}\:\:\Gamma\left({x}+\mathrm{1}\right) \\ $$$${grateful}... \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com