n-0-2n-1-8-n-n-2-Help-please-

Question Number 150432 by Jamshidbek last updated on 12/Aug/21

$$\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{2n}+\mathrm{1}\right)!}{\mathrm{8}^{\mathrm{n}} \centerdot\left(\mathrm{n}!\right)^{\mathrm{2}} }=?\:\:\:\:\:\mathrm{Help}\:\mathrm{please} \\ $$

Answered by Olaf_Thorendsen last updated on 12/Aug/21

f(x) = (1/( (1−x)^(3/2) )) = (1−x)^(−(3/2)) f′(x) = ((1.3)/2)(1−x)^(−(5/2)) f′′(x) = ((1.3.5)/2^2 )(1−x)^(−(7/2) ) f′′′(x) = ((1.3.5.7)/2^3 )(1−x)^(−(9/2) ) ... f^((n)) (x) = ((1.3.5.7...(2n+1))/2^n )(1−x)^(−(((2n+3))/2)) f^((n)) (x) = (((2n+1)!)/(2^n .2.4.6...2n))(1−x)^(−(((2n+3))/2)) f^((n)) (x) = (((2n+1)!)/(2^(2n) n!))(1−x)^(−(((2n+3))/2)) f^((n)) (0) = (((2n+1)!)/(2^(2n) n!)) f(x) = Σ_(n=0) ^∞ ((f^((n)) (0))/(n!))x^n f((1/2)) = (1/((1−(1/2))^(3/2) )) = Σ_(n=0) ^∞ (((2n+1)!)/(2^(2n) n!^2 )).((1/2))^n ⇒ Σ_(n=0) ^∞ (((2n+1)!)/(2^(3n) n!^2 )) = Σ_(n=0) ^∞ (((2n+1)!)/(8^n .n!^2 )) = 2(√2)

$${f}\left({x}\right)\:=\:\frac{\mathrm{1}}{\:\left(\mathrm{1}−{x}\right)^{\mathrm{3}/\mathrm{2}} }\:=\:\left(\mathrm{1}−{x}\right)^{−\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$${f}'\left({x}\right)\:=\:\frac{\mathrm{1}.\mathrm{3}}{\mathrm{2}}\left(\mathrm{1}−{x}\right)^{−\frac{\mathrm{5}}{\mathrm{2}}} \\ $$$${f}''\left({x}\right)\:=\:\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}}{\mathrm{2}^{\mathrm{2}} }\left(\mathrm{1}−{x}\right)^{−\frac{\mathrm{7}}{\mathrm{2}}\:} \\ $$$${f}'''\left({x}\right)\:=\:\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}}{\mathrm{2}^{\mathrm{3}} }\left(\mathrm{1}−{x}\right)^{−\frac{\mathrm{9}}{\mathrm{2}}\:\:} \\ $$$$… \\ $$$${f}^{\left({n}\right)} \left({x}\right)\:=\:\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}…\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{2}^{{n}} }\left(\mathrm{1}−{x}\right)^{−\frac{\left(\mathrm{2}{n}+\mathrm{3}\right)}{\mathrm{2}}} \\ $$$${f}^{\left({n}\right)} \left({x}\right)\:=\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{\mathrm{2}^{{n}} .\mathrm{2}.\mathrm{4}.\mathrm{6}…\mathrm{2}{n}}\left(\mathrm{1}−{x}\right)^{−\frac{\left(\mathrm{2}{n}+\mathrm{3}\right)}{\mathrm{2}}} \\ $$$${f}^{\left({n}\right)} \left({x}\right)\:=\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{\mathrm{2}^{\mathrm{2}{n}} {n}!}\left(\mathrm{1}−{x}\right)^{−\frac{\left(\mathrm{2}{n}+\mathrm{3}\right)}{\mathrm{2}}} \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{\mathrm{2}^{\mathrm{2}{n}} {n}!} \\ $$$${f}\left({x}\right)\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}{x}^{{n}} \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\:\frac{\mathrm{1}}{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}/\mathrm{2}} }\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{\mathrm{2}^{\mathrm{2}{n}} {n}!^{\mathrm{2}} }.\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} \\ $$$$\Rightarrow\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{\mathrm{2}^{\mathrm{3}{n}} {n}!^{\mathrm{2}} }\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{\mathrm{8}^{{n}} .{n}!^{\mathrm{2}} }\:=\:\mathrm{2}\sqrt{\mathrm{2}} \\ $$

Commented by Tawa11 last updated on 12/Aug/21