Question Number 142970 by mnjuly1970 last updated on 08/Jun/21
![..........CALCULUS........... prove that:: π:=Ξ£_(n=1) ^β (((β1)^(nβ1) ((nβ1)!)^2 )/((2n)!))=2log^2 (Ο) Ο=golden ratio.... .............](https://www.tinkutara.com/question/Q142970.png)
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:……….{CALCULUS}……….. \\ $$$$\:\:\:\:\:\:\:{prove}\:{that}::\:\: \\ $$$$\:\:\:\:\:\:\boldsymbol{\phi}:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(β\mathrm{1}\right)^{{n}β\mathrm{1}} \left(\left({n}β\mathrm{1}\right)!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}\right)!}=\mathrm{2}{log}^{\mathrm{2}} \left(\varphi\right) \\ $$$$\:\:\:\:\varphi={golden}\:{ratio}…. \\ $$$$\:\:\:\:…………. \\ $$
Answered by Dwaipayan Shikari last updated on 08/Jun/21
![Ξ£_(n=1) ^β ((((nβ1)!)^2 )/((2n)!))(β1)^(nβ1) =Ξ£_(n=1) ^β ((Ξ(n)Ξ(n))/(2nΞ(2n)))(β1)^(nβ1) =(1/2)β«_0 ^1 Ξ£_(n=1) ^β (β1)^(n+1) (1/n)x^(nβ1) (1βx)^(nβ1) dx =(1/2)β«_0 ^1 (1/(x(1βx)))log(1+x(1βx))dx =(1/2)β«_0 ^1 ((log(1+x(1βx)))/x)+((log(1+x(1βx)))/(1βx)) dx =β«_0 ^1 ((log(1+xβx^2 ))/x)dx =β«_0 ^1 ((log(x))/(xβ(1/Ο)))+((log(x))/(x+Ο))dx=β(1/Ο)Ξ£_(n=1) ^β β«_0 ^1 (xΟ)^n log(x)+(1/Ο)β«_0 ^1 Ξ£_(n=1) ^β (βxΟ)^n log(x)dx =(1/Ο){Ξ£_(n=1) ^β (Ο^n /((n+1)^2 ))+(1/Ο)Ξ£_(n=1) ^β (((βΟ)^n )/((n+1)^2 ))} =(1/Ο^2 )(Li_2 (Ο)βΟ)β(1/Ο^2 )(Li_2 (βΟ)+Ο) =(1/Ο^2 )(Li_2 (Ο)βLi_2 (βΟ))](https://www.tinkutara.com/question/Q142982.png)
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(\left({n}β\mathrm{1}\right)!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}\right)!}\left(β\mathrm{1}\right)^{{n}β\mathrm{1}} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\Gamma\left({n}\right)\Gamma\left({n}\right)}{\mathrm{2}{n}\Gamma\left(\mathrm{2}{n}\right)}\left(β\mathrm{1}\right)^{{n}β\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(β\mathrm{1}\right)^{{n}+\mathrm{1}} \frac{\mathrm{1}}{{n}}{x}^{{n}β\mathrm{1}} \left(\mathrm{1}β{x}\right)^{{n}β\mathrm{1}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{x}\left(\mathrm{1}β{x}\right)}{log}\left(\mathrm{1}+{x}\left(\mathrm{1}β{x}\right)\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}\left(\mathrm{1}+{x}\left(\mathrm{1}β{x}\right)\right)}{{x}}+\frac{{log}\left(\mathrm{1}+{x}\left(\mathrm{1}β{x}\right)\right)}{\mathrm{1}β{x}}\:{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}\left(\mathrm{1}+{x}β{x}^{\mathrm{2}} \right)}{{x}}{dx}\:\:\:\:\: \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}\left({x}\right)}{{x}β\frac{\mathrm{1}}{\varphi}}+\frac{{log}\left({x}\right)}{{x}+\varphi}{dx}=β\frac{\mathrm{1}}{\varphi}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}\varphi\right)^{{n}} {log}\left({x}\right)+\frac{\mathrm{1}}{\varphi}\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(β{x}\varphi\right)^{{n}} {log}\left({x}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\varphi}\left\{\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\varphi^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\varphi}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(β\varphi\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\right\} \\ $$$$=\frac{\mathrm{1}}{\varphi^{\mathrm{2}} }\left({Li}_{\mathrm{2}} \left(\varphi\right)β\varphi\right)β\frac{\mathrm{1}}{\varphi^{\mathrm{2}} }\left({Li}_{\mathrm{2}} \left(β\varphi\right)+\varphi\right) \\ $$$$=\frac{\mathrm{1}}{\varphi^{\mathrm{2}} }\left({Li}_{\mathrm{2}} \left(\varphi\right)β{Li}_{\mathrm{2}} \left(β\varphi\right)\right) \\ $$