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Question Number 119442 by mnjuly1970 last updated on 24/Oct/20
... advanced calculus... prove that : Σ_(n=1) ^∞ (1/(n^2 (((2n)),(n) ))) =^(???) ((ζ(2))/3) solution:: Σ_(n=1) ^∞ (1/(n^2 ∗(((2n)!)/((n!)^2 ))))=Σ_(n=1) ^∞ ((n!∗n!)/(n^2 ∗(2n)!)) =Σ_(n=1) ^∞ ((Γ(n)Γ(n+1))/(nΓ(2n+1)))=Σ_(n=1) ^∞ ((β(n,n+1))/n) =Σ_(n=1) ^∞ (1/n)∫_0 ^( 1) x^(n−1) (1−x)^n =∫_0 ^( 1) (1/x)Σ(((x−x^2 )^n )/n)dx =−∫_0 ^( 1) ((ln(1−x+x^2 ))/x)dx =−∫_0 ^( 1) ((ln(1+x^3 )−ln(1+x))/x)dx =∫_0 ^( 1) ((ln(1+x))/x)dx −∫_0 ^( 1) ((ln(1+x^3 ))/x)dx =−li_2 (−1) −∫_0 ^( 1) ((Σ_(n=1) (((−1)^(n+1) x^(3n) )/n))/x) dx =(π^2 /(12))+Σ_(n=1) ^∞ (((−1)^n )/n)∫_0 ^( 1) x^(3n−1) dx =(π^2 /(12)) +(1/3)Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =(π^2 /(12))−(π^2 /(36)) =(π^2 /(18)) =((ζ(2))/3) ✓✓ m.n.july.1970..
$$\:\:\:\:\:\:\:\:…\:{advanced}\:\:{calculus}… \\ $$$$\:\:\:\:\:\:{prove}\:\:{that}\::\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} \begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}}\:\overset{???} {=}\frac{\zeta\left(\mathrm{2}\right)}{\mathrm{3}} \\ $$$${solution}::\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} \ast\frac{\left(\mathrm{2}{n}\right)!}{\left({n}!\right)^{\mathrm{2}} }}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}!\ast{n}!}{{n}^{\mathrm{2}} \ast\left(\mathrm{2}{n}\right)!}\: \\ $$$$\:\:\:\:\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\Gamma\left({n}\right)\Gamma\left({n}+\mathrm{1}\right)}{{n}\Gamma\left(\mathrm{2}{n}+\mathrm{1}\right)}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\beta\left({n},{n}+\mathrm{1}\right)}{{n}} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{{n}−\mathrm{1}} \left(\mathrm{1}−{x}\right)^{{n}} \\ $$$$=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}}{{x}}\Sigma\frac{\left({x}−{x}^{\mathrm{2}} \right)^{{n}} }{{n}}{dx} \\ $$$$=−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right)}{{x}}{dx} \\ $$$$\:\:=−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)−{ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx} \\ $$$$\:\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx}\:−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)}{{x}}{dx} \\ $$$$\:=−{li}_{\mathrm{2}} \left(−\mathrm{1}\right)\:−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\underset{{n}=\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {x}^{\mathrm{3}{n}} }{{n}}}{{x}}\:{dx} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{\mathrm{3}{n}−\mathrm{1}} {dx}\:\:\:\:\: \\ $$$$ \\ $$$$\:=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:+\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\frac{\pi^{\mathrm{2}} }{\mathrm{36}} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{18}}\:=\frac{\zeta\left(\mathrm{2}\right)}{\mathrm{3}}\:\:\checkmark\checkmark \\ $$$$\:\:\:\:\:\:\:\:{m}.{n}.{july}.\mathrm{1970}.. \\ $$$$\:\: \\ $$$$ \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$

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