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Question Number 116162 by mnjuly1970 last updated on 01/Oct/20

$$$$$$$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{2n+1}{{16}^n(n^2+3n+2)}{\left(\begin{array}{c}2n\\ n\end{array}\right)}^2=\frac{8}{3\pi }$$$$$$$$m.n.july1970.$$$$$$

Answered by maths mind last updated on 02/Oct/20

$$wetrytoprouvethat$$$$S\left(n\right)\underset{k=0}{\sum ^n}\frac{2k+1}{{16}^k(k^2+3k+2)}{\left(\begin{array}{c}2k\\ k\end{array}\right)}^2=\frac{(n+1)(2n+3){(2(n+1)!)}^2}{4^{2n+1}.3(n+2){((n+1)!)}^4}=f\left(n\right)?$$$$k^2+3k+2=(k+2)(k+1)$$$$S\left(0\right)=\frac{1}{2}=f\left(0\right)$$$$suppose\mathrm{\forall }n\in \mathbb{N}S\left(n\right)=f\left(n\right)weShowthats(n+1)=f(n+1)$$$$S(n+1)=S\left(n\right)+\frac{2(n+1)+1}{{16}^{n+1}(n+3)(n+2)}.{\left(\begin{array}{c}2n+2\\ n+1\end{array}\right)}^2$$$$s\left(n\right)=f\left(n\right)=\frac{(n+1)(2n+3){(2(n+1)!)}^2}{4^{2n+1}.3(n+2){((n+1)!)}^4}$$$$S\left(n\right)=f\left(n\right)+\frac{2n+3}{{16}^{n+1}(n+3)(n+2)}{\left(\begin{array}{c}2n+2\\ n+1\end{array}\right)}^2$$$$=\frac{(n+1)(2n+3){(2(n+1)!)}^2}{4^{2n+1}3(n+2){((n+1)!)}^4}+\frac{2n+3}{{16}^{n+1}(n+3)(n+2)}.\frac{{((2n+2)!)}^2}{{((n+1)!)}^4}$$$$=\frac{(2n+3){((2n+2)!)}^2}{4^{2n+1}{((n+1)!)}^4(n+2)}(\frac{n+1}{3}+\frac{1}{4(n+3)})$$$$=\frac{(2n+3){((2n+2)!)}^2}{4^{2n+1}{((n+1)!)}^4(n+2)}\left(\frac{4(n+1)(n+3)+3}{12}\right)$$$$=\frac{(2n+3)!.(2n+2)!}{4^{2n+1}(n+2)!{((n+1)!)}^3}\left(\frac{4n^2+16n+15}{12(n+3)}\right)$$$$\frac{-16-4}{8}=-\frac{5}{2},\frac{-12}{8}=-\frac{3}{2}$$$$=\frac{(2n+3)!(2n+2)!\left(4(n+\frac{5}{2})(n+\frac{3}{2})\right)}{4^{2n+1}(n+2)!{((n+1)!)}^3.12(n+3)}=\frac{(2n+5){((2n+3)!)}^2}{4^{2n+1}.12.{((n+1)!)}^3(n+2)!(n+3)}$$$$=\frac{(2n+5){((2n+3)!)}^2.(2n+4)(2n+4).(n+2)}{4^{2n+2}.3(n+3)((n+2)!){((n+1)!)}^3.(2n+4)(2n+4)(n+2)}$$$$=\frac{(2n+5)(n+2){((2n+4)!)}^2}{4^{2n+2}.3(n+3)(n+2)!.{((n+1)!)}^3.4(n+2)(n+2)(n+2)}$$$$=\frac{(n+2)(2n+5){((2n+4)!)}^2}{4^{2n+3}.3{((n+2)!)}^4(n+3)}=\frac{(n+1+1)(2(n+1)+3){(2(n+1+1)!)}^2}{4^{2(n+1)+1}.3{((n+1+1)!)}^4(n+1+2)}$$$$=f(n+1)$$$$\Rightarrow \mathrm{\forall }n\in \mathbb{N}\underset{k=0}{\sum ^n}\frac{2k+1}{{16}^k(k^2+3k+2)}.{\left(\begin{array}{c}2k\\ k\end{array}\right)}^2=\frac{(n+1)(2n+3){((2n+2)!)}^2}{4^{2n+1}.3{((n+1)!)}^4(n+2)}$$$$\sum _{k⩾0}\frac{2k+1}{{16}^k(k^2+3k+2)}{\left(\begin{array}{c}2k\\ k\end{array}\right)}^2=\underset{n\to \mathrm{\infty }}{\lim }f\left(n\right)$$$$Strilingformula\Rightarrow$$$$n!\sim \sqrt{2\pi n}{\left(\frac{n}{e}\right)}^n$$$$\Rightarrow \underset{n\to \mathrm{\infty }}{\lim }\frac{(n+1)(2n+3){((2n+2)!)}^2}{4^{2n+1}.3{((n+1)!)}^4(n+2)}=\underset{n\to \mathrm{\infty }}{\lim }\frac{(n+1)(2n+3){(\sqrt{2\pi (2n+2)}.\frac{{(2n+2)}^{2n+2}}{e^{2n+2}})}^2}{4^{2n+1}(n+2).3.{(\sqrt{2\pi (n+1)}.{\left(\frac{n+1}{e}\right)}^{n+1})}^4}$$$$=\underset{n\to \mathrm{\infty }}{\lim }\frac{(n+1)(2n+3).2\pi (2n+2).\frac{{(2n+2)}^{4n+4}}{e^{4n+4}}}{4^{2n+1}(n+2).3{\left(2\pi (n+1)\right)}^2.{\left(\frac{n+1}{e}\right)}^{4n+4}}$$$$=\underset{n\to \mathrm{\infty }}{\lim }\frac{(n+1)(2n+3)(2n+2).2\pi }{4^{2n+1}(n+2).3.4{\pi }^2{(n+1)}^2}.\frac{2^{4n+4}{(n+1)}^{4n+4}}{e^{4n+4}}.\frac{e^{4n+4}}{{(n+1)}^{4n+4}}$$$$=\underset{n\to \mathrm{\infty }}{\lim }\frac{(2n+3)(2n+2)(n+1).4^{2n+2}}{4^{2n+1}.6\pi (n+1)(n+1)(n+2)}$$$$=\underset{n\to \mathrm{\infty }}{\lim }\frac{4}{.3.2\pi }.\frac{2n+3}{n+1}.\frac{2n+2}{n+1}.\frac{n+1}{n+2}=\frac{4}{6\pi }.2.2=\frac{16}{6\pi }=\frac{8}{3\pi }$$$$\sum _{n⩾0}\frac{2n+1}{{16}^n(n^2+3n+2)}{\left(\begin{array}{c}2n\\ n\end{array}\right)}^2=\frac{8}{3\pi }$$$$$$$$$$$$$$$$$$$$$$

Commented by mnjuly1970 last updated on 02/Oct/20

$$veryastonishing$$$$thankyousir...excellent..$$

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