Question Number 144322 by qaz last updated on 24/Jun/21
$$\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{2n}\right)!!}{\left(\mathrm{2n}+\mathrm{1}\right)!!\left(\mathrm{n}+\mathrm{1}\right)}\mathrm{x}^{\mathrm{2n}+\mathrm{2}} =?……….\mid\mathrm{x}\mid\leqslant\mathrm{1} \\ $$
Answered by mindispower last updated on 25/Jun/21
$$\left(\mathrm{2}{n}\right)!!=\mathrm{2}^{{n}} .{n}! \\ $$$$\left(\mathrm{2}{n}+\mathrm{1}\right)!!=\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{\mathrm{2}^{{n}} .{n}!} \\ $$$$\Leftrightarrow\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{2}^{{n}} .{n}!}{\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{\mathrm{2}^{{n}} {n}!}\left({n}+\mathrm{1}\right)}{x}^{\mathrm{2}{n}+\mathrm{2}} \\ $$$$=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{2}^{\mathrm{2}{n}} .\left({n}!\right)^{\mathrm{2}} {x}^{\mathrm{2}{n}+\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!.\left({n}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{2}^{\mathrm{2}\left({n}+\mathrm{1}\right)} .\left({n}!\left({n}+\mathrm{1}\right)\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{2}\right)!\left({n}+\mathrm{1}\right)^{\mathrm{2}} }.{x}^{\mathrm{2}\left({n}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}.\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(\mathrm{2}^{{n}} \right)^{\mathrm{2}} .\left({n}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}\right)!{n}^{\mathrm{2}} }.{x}^{\mathrm{2}{n}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(\mathrm{2}{x}\right)^{\mathrm{2}{n}} }{{n}^{\mathrm{2}} .{C}_{\mathrm{2}{n}} ^{{n}} }={arcsin}^{\mathrm{2}} \left({x}\right) \\ $$$$ \\ $$$$ \\ $$$$\: \\ $$