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Question Number 144322 by qaz last updated on 24/Jun/21

Σ_(n=0) ^∞ (((2n)!!)/((2n+1)!!(n+1)))x^(2n+2) =?..........∣x∣≤1

∑∞n=0(2n)!!(2n+1)!!(n+1)x2n+2=?..........∣x∣⩽1

Answered by mindispower last updated on 25/Jun/21

(2n)!!=2^n .n! (2n+1)!!=(((2n+1)!)/(2^n .n!)) ⇔Σ_(n≥0) ((2^n .n!)/((((2n+1)!)/(2^n n!))(n+1)))x^(2n+2) =Σ_(n≥0) ((2^(2n) .(n!)^2 x^(2n+2) )/((2n+1)!.(n+1))) =(1/2)Σ_(n≥0) ((2^(2(n+1)) .(n!(n+1))^2 )/((2n+2)!(n+1)^2 )).x^(2(n+1)) =(1/2).Σ_(n≥1) (((2^n )^2 .(n!)^2 )/((2n)!n^2 )).x^(2n) =(1/2)Σ_(n≥1) (((2x)^(2n) )/(n^2 .C_(2n) ^n ))=arcsin^2 (x)

(2n)!!=2n.n!(2n+1)!!=(2n+1)!2n.n!⇔∑n⩾02n.n!(2n+1)!2nn!(n+1)x2n+2=∑n⩾022n.(n!)2x2n+2(2n+1)!.(n+1)=12∑n⩾022(n+1).(n!(n+1))2(2n+2)!(n+1)2.x2(n+1)=12.∑n⩾1(2n)2.(n!)2(2n)!n2.x2n=12∑n⩾1(2x)2nn2.C2nn=arcsin2(x)

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