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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

$$\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{2n}\right)!!}{\left(\mathrm{2n}+\mathrm{1}\right)!!\left(\mathrm{n}+\mathrm{1}\right)}\mathrm{x}^{\mathrm{2n}+\mathrm{2}} =?..........\mid\mathrm{x}\mid\leqslant\mathrm{1} \\ $$

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)

$$\left(\mathrm{2}{n}\right)!!=\mathrm{2}^{{n}} .{n}! \\ $$$$\left(\mathrm{2}{n}+\mathrm{1}\right)!!=\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{\mathrm{2}^{{n}} .{n}!} \\ $$$$\Leftrightarrow\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{2}^{{n}} .{n}!}{\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{\mathrm{2}^{{n}} {n}!}\left({n}+\mathrm{1}\right)}{x}^{\mathrm{2}{n}+\mathrm{2}} \\ $$$$=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{2}^{\mathrm{2}{n}} .\left({n}!\right)^{\mathrm{2}} {x}^{\mathrm{2}{n}+\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!.\left({n}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{2}^{\mathrm{2}\left({n}+\mathrm{1}\right)} .\left({n}!\left({n}+\mathrm{1}\right)\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{2}\right)!\left({n}+\mathrm{1}\right)^{\mathrm{2}} }.{x}^{\mathrm{2}\left({n}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}.\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(\mathrm{2}^{{n}} \right)^{\mathrm{2}} .\left({n}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}\right)!{n}^{\mathrm{2}} }.{x}^{\mathrm{2}{n}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(\mathrm{2}{x}\right)^{\mathrm{2}{n}} }{{n}^{\mathrm{2}} .{C}_{\mathrm{2}{n}} ^{{n}} }={arcsin}^{\mathrm{2}} \left({x}\right) \\ $$$$ \\ $$$$ \\ $$$$\: \\ $$

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