Show-that-n-1-n-2-2n-1-3-2pi-3-27-

Question Number 197935 by Frix last updated on 04/Oct/23

Show that Σ_(n=1) ^∞ (((n!)^2 )/((2n)!)) =(1/3)+((2π(√3))/(27))

$$\mathrm{Show}\:\mathrm{that} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\left({n}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}\right)!}\:=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{2}\pi\sqrt{\mathrm{3}}}{\mathrm{27}} \\ $$

Answered by witcher3 last updated on 05/Oct/23

Σ_(n≥1) ((Γ(n+1)Γ(n).n)/(Γ(2n+1))) =Σ_(n≥1) nβ(n,n+1)=Σ_(n≥1) ∫_0 ^1 nt^n (1−t)^(n−1) dt Σ_(n≥1) nx^(n−1) =(1/((1−x)^2 )) =∫_0 ^1 (x/((1−x(1−x))^2 ))dx=∫_0 ^1 ((1−x)/((1−x(1−x))^2 ))dx=A A=(1/2)∫_0 ^1 (dx/(((x−(1/2))^2 +(3/4)))) =(1/2)∫_(−(1/2)) ^(1/2) (dx/((x^2 +(3/4))^2 )),∫_(−(1/2)) ^(1/2) (dx/((x^2 +a)))=f(a) f′(a)=∫((−dx)/((x^2 +a)^2 ))... f(a)=(2/( (√a)))tan^(−1) ((1/( 2(√a)))) f′(a)=−(a)^(−(3/2)) tan^(−1) ((1/(2(√a))))+(2/( (√a))).(−(1/4)a^(−(3/2)) .((4a)/(4a+1)).) A=−(1/2).f′((3/4)) =(1/2)((8/(3(√3))).(π/6))+(2/( (√3)))((1/4).(8/(3(√3))).(3/4)) =((2π)/(9(√3)))+(1/3)=((2π(√3))/(27))+(1/3) ⇔Σ_(n≥1) (((n!)^2 )/(2n))=((2π(√3))/(27))+(1/3)

$$\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\Gamma\left(\mathrm{n}+\mathrm{1}\right)\Gamma\left(\mathrm{n}\right).\mathrm{n}}{\Gamma\left(\mathrm{2n}+\mathrm{1}\right)} \\ $$$$=\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\mathrm{n}\beta\left(\mathrm{n},\mathrm{n}+\mathrm{1}\right)=\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{nt}^{\mathrm{n}} \left(\mathrm{1}−\mathrm{t}\right)^{\mathrm{n}−\mathrm{1}} \mathrm{dt} \\ $$$$\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\mathrm{nx}^{\mathrm{n}−\mathrm{1}} =\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}}{\left(\mathrm{1}−\mathrm{x}\left(\mathrm{1}−\mathrm{x}\right)\right)^{\mathrm{2}} }\mathrm{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−\mathrm{x}}{\left(\mathrm{1}−\mathrm{x}\left(\mathrm{1}−\mathrm{x}\right)\right)^{\mathrm{2}} }\mathrm{dx}=\mathrm{A} \\ $$$$\mathrm{A}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dx}}{\left(\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} },\int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{a}\right)}=\mathrm{f}\left(\mathrm{a}\right) \\ $$$$\mathrm{f}'\left(\mathrm{a}\right)=\int\frac{−\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{a}\right)^{\mathrm{2}} }… \\ $$$$\mathrm{f}\left(\mathrm{a}\right)=\frac{\mathrm{2}}{\:\sqrt{\mathrm{a}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\mathrm{2}\sqrt{\mathrm{a}}}\right) \\ $$$$\mathrm{f}'\left(\mathrm{a}\right)=−\left(\mathrm{a}\right)^{−\frac{\mathrm{3}}{\mathrm{2}}} \mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{a}}}\right)+\frac{\mathrm{2}}{\:\sqrt{\mathrm{a}}}.\left(−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{a}^{−\frac{\mathrm{3}}{\mathrm{2}}} .\frac{\mathrm{4a}}{\mathrm{4a}+\mathrm{1}}.\right) \\ $$$$\mathrm{A}=−\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{f}'\left(\frac{\mathrm{3}}{\mathrm{4}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{8}}{\mathrm{3}\sqrt{\mathrm{3}}}.\frac{\pi}{\mathrm{6}}\right)+\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\left(\frac{\mathrm{1}}{\mathrm{4}}.\frac{\mathrm{8}}{\mathrm{3}\sqrt{\mathrm{3}}}.\frac{\mathrm{3}}{\mathrm{4}}\right) \\ $$$$=\frac{\mathrm{2}\pi}{\mathrm{9}\sqrt{\mathrm{3}}}+\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{2}\pi\sqrt{\mathrm{3}}}{\mathrm{27}}+\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Leftrightarrow\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\left(\mathrm{n}!\right)^{\mathrm{2}} }{\mathrm{2n}}=\frac{\mathrm{2}\pi\sqrt{\mathrm{3}}}{\mathrm{27}}+\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by witcher3 last updated on 05/Oct/23

$$\mathrm{y}'\mathrm{re}\:\mathrm{Welcom} \\ $$

Commented by Frix last updated on 05/Oct/23

$$\mathrm{Thank}\:\mathrm{you}! \\ $$

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