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n-1-1-n-4n-2n-




Question Number 121985 by Dwaipayan Shikari last updated on 13/Nov/20
Σ_(n=1) ^∞ (1/(n (((4n)),((2n)) )))
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{n}\begin{pmatrix}{\mathrm{4n}}\\{\mathrm{2n}}\end{pmatrix}} \\ $$
Commented by Dwaipayan Shikari last updated on 13/Nov/20
Σ_(n=1) ^∞ (((2n!)^2 )/(n(4n!)))  =Σ_(n=1) ^∞ ((Γ(2n+1)Γ(2n+1))/(nΓ(2n+1)))  =2Σ_(n=1) ^∞ ((nΓ(2n)Γ(2n+1))/(nΓ(4n+1)))  =2Σ_(n=1) ^∞ β(2n,2n+1)  =2Σ_(n=1) ^∞ ∫_0 ^1 x^(2n−1) (1−x)^(2n) dx  =2∫_0 ^1 Σ_(n=1) ^∞ x^(2n−1) (1−x)^(2n)  dx              S=x(1−x)^2 +x^3 (1−x)^4 +...  −Sx^2 (1−x)^2 =−x^3 (1−x)^4 −x^5 (1−x)^6 −...  S(1−x^2 +2x^3 −x^4 )=x(1−x)^2   S=((x(1−x)^2 )/(1−x^2 +2x^3 −x^4 ))  So  =2∫_0 ^1 ((x(1−x)^2 )/(1−x^2 +2x^3 −x^4 ))dx  =−2∫_0 ^1 ((−x^3 +2x^2 −x)/(1−x^2 +2x^3 −x^4 ))dx  =−2∫_0 ^1 ((−2x^3 +3x^2 −x)/(1−x^2 +2x^3 −x^4 ))+((x^3 −x^2 )/(1−x^2 +2x^3 −x^4 ))dx  =−∫_0 ^1 ((−4x^3 +6x^2 −2x)/(1−x^2 +2x^3 −x^4 ))+2∫((x^2 −x^3 )/(1−x^2 +2x^3 −x^4 ))dx  =−[log(1−x^2 +2x^3 −x^4 )]_0 ^1 +2∫((x^2 (1−x))/(1−x^2 (1−x)^2 ))dx
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{2}{n}!\right)^{\mathrm{2}} }{{n}\left(\mathrm{4}{n}!\right)} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\Gamma\left(\mathrm{2}{n}+\mathrm{1}\right)\Gamma\left(\mathrm{2}{n}+\mathrm{1}\right)}{{n}\Gamma\left(\mathrm{2}{n}+\mathrm{1}\right)} \\ $$$$=\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}\Gamma\left(\mathrm{2}{n}\right)\Gamma\left(\mathrm{2}{n}+\mathrm{1}\right)}{{n}\Gamma\left(\mathrm{4}{n}+\mathrm{1}\right)} \\ $$$$=\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\beta\left(\mathrm{2}{n},\mathrm{2}{n}+\mathrm{1}\right) \\ $$$$=\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{n}−\mathrm{1}} \left(\mathrm{1}−{x}\right)^{\mathrm{2}{n}} {dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{x}^{\mathrm{2}{n}−\mathrm{1}} \left(\mathrm{1}−{x}\right)^{\mathrm{2}{n}} \:{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{S}={x}\left(\mathrm{1}−{x}\right)^{\mathrm{2}} +{x}^{\mathrm{3}} \left(\mathrm{1}−{x}\right)^{\mathrm{4}} +… \\ $$$$−{Sx}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)^{\mathrm{2}} =−{x}^{\mathrm{3}} \left(\mathrm{1}−{x}\right)^{\mathrm{4}} −{x}^{\mathrm{5}} \left(\mathrm{1}−{x}\right)^{\mathrm{6}} −… \\ $$$${S}\left(\mathrm{1}−{x}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{3}} −{x}^{\mathrm{4}} \right)={x}\left(\mathrm{1}−{x}\right)^{\mathrm{2}} \\ $$$${S}=\frac{{x}\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{3}} −{x}^{\mathrm{4}} } \\ $$$${So} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{3}} −{x}^{\mathrm{4}} }{dx} \\ $$$$=−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{−{x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} −{x}}{\mathrm{1}−{x}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{3}} −{x}^{\mathrm{4}} }{dx} \\ $$$$=−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{−\mathrm{2}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} −{x}}{\mathrm{1}−{x}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{3}} −{x}^{\mathrm{4}} }+\frac{{x}^{\mathrm{3}} −{x}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{3}} −{x}^{\mathrm{4}} }{dx} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{−\mathrm{4}{x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} −\mathrm{2}{x}}{\mathrm{1}−{x}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{3}} −{x}^{\mathrm{4}} }+\mathrm{2}\int\frac{{x}^{\mathrm{2}} −{x}^{\mathrm{3}} }{\mathrm{1}−{x}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{3}} −{x}^{\mathrm{4}} }{dx} \\ $$$$=−\left[{log}\left(\mathrm{1}−{x}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{3}} −{x}^{\mathrm{4}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\mathrm{2}\int\frac{{x}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)}{\mathrm{1}−{x}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)^{\mathrm{2}} }{dx} \\ $$

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