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

Σ_(n=0) ^∞ ((2n+1)/(16^n (n^2 +3n+2))) (((2n)),(n) )^2 =(8/(3π)) m.n.july 1970.

$$\:\: \\ $$$$ \\ $$$$\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{16}^{{n}} \left({n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}\right)}\:\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}^{\mathrm{2}} \:=\frac{\mathrm{8}}{\mathrm{3}\pi}\: \\ $$$$ \\ $$$${m}.{n}.{july}\:\mathrm{1970}. \\ $$$$\: \\ $$

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

we try to prouve that S(n)Σ_(k=0) ^n ((2k+1)/(16^k (k^2 +3k+2))) (((2k)),(k) )^2 =(((n+1)(2n+3)(2(n+1)!)^2 )/(4^(2n+1) .3(n+2)((n+1)!)^4 ))=f(n)? k^2 +3k+2=(k+2)(k+1) S(0)=(1/2)=f(0) suppose ∀n∈N S(n)=f(n) we Show that s(n+1)=f(n+1) S(n+1)=S(n)+((2(n+1)+1)/(16^(n+1) (n+3)(n+2))). (((2n+2)),((n+1)) )^2 s(n)=f(n)=(((n+1)(2n+3)(2(n+1)!)^2 )/(4^(2n+1) .3(n+2)((n+1)!)^4 )) S(n)=f(n)+((2n+3)/(16^(n+1) (n+3)(n+2))) (((2n+2)),((n+1)) )^2 =(((n+1)(2n+3)(2(n+1)!)^2 )/(4^(2n+1) 3(n+2)((n+1)!)^4 ))+((2n+3)/(16^(n+1) (n+3)(n+2))).((((2n+2)!)^2 )/(((n+1)!)^4 )) =(((2n+3)((2n+2)!)^2 )/(4^(2n+1) ((n+1)!)^4 (n+2)))(((n+1)/3)+(1/(4(n+3)))) =(((2n+3)((2n+2)!)^2 )/(4^(2n+1) ((n+1)!)^4 (n+2)))(((4(n+1)(n+3)+3)/(12))) =(((2n+3)!.(2n+2)!)/(4^(2n+1) (n+2)!((n+1)!)^3 ))(((4n^2 +16n+15)/(12(n+3)))) ((−16−4)/8)=−(5/2),((−12)/8)=−(3/2) =(((2n+3)!(2n+2)!(4(n+(5/2))(n+(3/2))))/(4^(2n+1) (n+2)!((n+1)!)^3 .12(n+3)))=(((2n+5)((2n+3)!)^2 )/(4^(2n+1) .12.((n+1)!)^3 (n+2)!(n+3))) =(((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))) =(((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))) =(((n+2)(2n+5)((2n+4)!)^2 )/(4^(2n+3) .3((n+2)!)^4 (n+3)))=(((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) ⇒∀n∈N Σ_(k=0) ^n ((2k+1)/(16^k (k^2 +3k+2))). (((2k)),(k) )^2 =(((n+1)(2n+3)((2n+2)!)^2 )/(4^(2n+1) .3((n+1)!)^4 (n+2))) Σ_(k≥0) ((2k+1)/(16^k (k^2 +3k+2))) (((2k)),(k) )^2 =lim_(n→∞) f(n) Striling formula⇒ n!∼(√(2πn))((n/e))^n ⇒lim_(n→∞) (((n+1)(2n+3)((2n+2)!)^2 )/(4^(2n+1) .3((n+1)!)^4 (n+2)))=lim_(n→∞) (((n+1)(2n+3)((√(2π(2n+2))).(((2n+2)^(2n+2) )/e^(2n+2) ))^2 )/(4^(2n+1) (n+2).3.((√(2π(n+1))).(((n+1)/e))^(n+1) )^4 )) =lim_(n→∞) (((n+1)(2n+3).2π(2n+2).(((2n+2)^(4n+4) )/e^(4n+4) ))/(4^(2n+1) (n+2).3(2π(n+1))^2 .(((n+1)/e))^(4n+4) )) =lim_(n→∞) (((n+1)(2n+3)(2n+2).2π)/(4^(2n+1) (n+2).3.4π^2 (n+1)^2 )).((2^(4n+4) (n+1)^(4n+4) )/e^(4n+4) ).(e^(4n+4) /((n+1)^(4n+4) )) =lim_(n→∞) (((2n+3)(2n+2)(n+1).4^(2n+2) )/(4^(2n+1) .6π(n+1)(n+1)(n+2))) =lim_(n→∞) (4/(.3.2π)).((2n+3)/(n+1)).((2n+2)/(n+1)).((n+1)/(n+2))=(4/(6π)).2.2=((16)/(6π))=(8/(3π)) Σ_(n≥0) ((2n+1)/(16^n (n^2 +3n+2))) (((2n)),(n) )^2 =(8/(3π))

$${we}\:{try}\:{to}\:{prouve}\:{that} \\ $$$${S}\left({n}\right)\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{16}^{{k}} \left({k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{2}\right)}\begin{pmatrix}{\mathrm{2}{k}}\\{{k}}\end{pmatrix}^{\mathrm{2}} =\frac{\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)\left(\mathrm{2}\left({n}+\mathrm{1}\right)!\right)^{\mathrm{2}} }{\mathrm{4}^{\mathrm{2}{n}+\mathrm{1}} .\mathrm{3}\left({n}+\mathrm{2}\right)\left(\left({n}+\mathrm{1}\right)!\right)^{\mathrm{4}} }={f}\left({n}\right)? \\ $$$${k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{2}=\left({k}+\mathrm{2}\right)\left({k}+\mathrm{1}\right) \\ $$$${S}\left(\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{2}}={f}\left(\mathrm{0}\right) \\ $$$${suppose}\:\forall{n}\in\mathbb{N}\:\:{S}\left({n}\right)={f}\left({n}\right)\:{we}\:{Show}\:{that}\:{s}\left({n}+\mathrm{1}\right)={f}\left({n}+\mathrm{1}\right) \\ $$$${S}\left({n}+\mathrm{1}\right)={S}\left({n}\right)+\frac{\mathrm{2}\left({n}+\mathrm{1}\right)+\mathrm{1}}{\mathrm{16}^{{n}+\mathrm{1}} \left({n}+\mathrm{3}\right)\left({n}+\mathrm{2}\right)}.\begin{pmatrix}{\mathrm{2}{n}+\mathrm{2}}\\{{n}+\mathrm{1}}\end{pmatrix}^{\mathrm{2}} \\ $$$${s}\left({n}\right)={f}\left({n}\right)=\frac{\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)\left(\mathrm{2}\left({n}+\mathrm{1}\right)!\right)^{\mathrm{2}} }{\mathrm{4}^{\mathrm{2}{n}+\mathrm{1}} .\mathrm{3}\left({n}+\mathrm{2}\right)\left(\left({n}+\mathrm{1}\right)!\right)^{\mathrm{4}} } \\ $$$${S}\left({n}\right)={f}\left({n}\right)+\frac{\mathrm{2}{n}+\mathrm{3}}{\mathrm{16}^{{n}+\mathrm{1}} \left({n}+\mathrm{3}\right)\left({n}+\mathrm{2}\right)}\begin{pmatrix}{\mathrm{2}{n}+\mathrm{2}}\\{{n}+\mathrm{1}}\end{pmatrix}^{\mathrm{2}} \\ $$$$=\frac{\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)\left(\mathrm{2}\left({n}+\mathrm{1}\right)!\right)^{\mathrm{2}} }{\mathrm{4}^{\mathrm{2}{n}+\mathrm{1}} \mathrm{3}\left({n}+\mathrm{2}\right)\left(\left({n}+\mathrm{1}\right)!\right)^{\mathrm{4}} }+\frac{\mathrm{2}{n}+\mathrm{3}}{\mathrm{16}^{{n}+\mathrm{1}} \left({n}+\mathrm{3}\right)\left({n}+\mathrm{2}\right)}.\frac{\left(\left(\mathrm{2}{n}+\mathrm{2}\right)!\right)^{\mathrm{2}} }{\left(\left({n}+\mathrm{1}\right)!\right)^{\mathrm{4}} } \\ $$$$=\frac{\left(\mathrm{2}{n}+\mathrm{3}\right)\left(\left(\mathrm{2}{n}+\mathrm{2}\right)!\right)^{\mathrm{2}} }{\mathrm{4}^{\mathrm{2}{n}+\mathrm{1}} \left(\left({n}+\mathrm{1}\right)!\right)^{\mathrm{4}} \left({n}+\mathrm{2}\right)}\left(\frac{{n}+\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}\left({n}+\mathrm{3}\right)}\right) \\ $$$$=\frac{\left(\mathrm{2}{n}+\mathrm{3}\right)\left(\left(\mathrm{2}{n}+\mathrm{2}\right)!\right)^{\mathrm{2}} }{\mathrm{4}^{\mathrm{2}{n}+\mathrm{1}} \left(\left({n}+\mathrm{1}\right)!\right)^{\mathrm{4}} \left({n}+\mathrm{2}\right)}\left(\frac{\mathrm{4}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{3}\right)+\mathrm{3}}{\mathrm{12}}\right) \\ $$$$=\frac{\left(\mathrm{2}{n}+\mathrm{3}\right)!.\left(\mathrm{2}{n}+\mathrm{2}\right)!}{\mathrm{4}^{\mathrm{2}{n}+\mathrm{1}} \left({n}+\mathrm{2}\right)!\left(\left({n}+\mathrm{1}\right)!\right)^{\mathrm{3}} }\left(\frac{\mathrm{4}{n}^{\mathrm{2}} +\mathrm{16}{n}+\mathrm{15}}{\mathrm{12}\left({n}+\mathrm{3}\right)}\right) \\ $$$$\frac{−\mathrm{16}−\mathrm{4}}{\mathrm{8}}=−\frac{\mathrm{5}}{\mathrm{2}},\frac{−\mathrm{12}}{\mathrm{8}}=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$=\frac{\left(\mathrm{2}{n}+\mathrm{3}\right)!\left(\mathrm{2}{n}+\mathrm{2}\right)!\left(\mathrm{4}\left({n}+\frac{\mathrm{5}}{\mathrm{2}}\right)\left({n}+\frac{\mathrm{3}}{\mathrm{2}}\right)\right)}{\mathrm{4}^{\mathrm{2}{n}+\mathrm{1}} \left({n}+\mathrm{2}\right)!\left(\left({n}+\mathrm{1}\right)!\right)^{\mathrm{3}} .\mathrm{12}\left({n}+\mathrm{3}\right)}=\frac{\left(\mathrm{2}{n}+\mathrm{5}\right)\left(\left(\mathrm{2}{n}+\mathrm{3}\right)!\right)^{\mathrm{2}} }{\mathrm{4}^{\mathrm{2}{n}+\mathrm{1}} .\mathrm{12}.\left(\left({n}+\mathrm{1}\right)!\right)^{\mathrm{3}} \left({n}+\mathrm{2}\right)!\left({n}+\mathrm{3}\right)} \\ $$$$=\frac{\left(\mathrm{2}{n}+\mathrm{5}\right)\left(\left(\mathrm{2}{n}+\mathrm{3}\right)!\right)^{\mathrm{2}} .\left(\mathrm{2}{n}+\mathrm{4}\right)\left(\mathrm{2}{n}+\mathrm{4}\right).\left({n}+\mathrm{2}\right)}{\mathrm{4}^{\mathrm{2}{n}+\mathrm{2}} .\mathrm{3}\left({n}+\mathrm{3}\right)\left(\left({n}+\mathrm{2}\right)!\right)\left(\left({n}+\mathrm{1}\right)!\right)^{\mathrm{3}} .\left(\mathrm{2}{n}+\mathrm{4}\right)\left(\mathrm{2}{n}+\mathrm{4}\right)\left({n}+\mathrm{2}\right)} \\ $$$$=\frac{\left(\mathrm{2}{n}+\mathrm{5}\right)\left({n}+\mathrm{2}\right)\left(\left(\mathrm{2}{n}+\mathrm{4}\right)!\right)^{\mathrm{2}} }{\mathrm{4}^{\mathrm{2}{n}+\mathrm{2}} .\mathrm{3}\left({n}+\mathrm{3}\right)\left({n}+\mathrm{2}\right)!.\left(\left({n}+\mathrm{1}\right)!\right)^{\mathrm{3}} .\mathrm{4}\left({n}+\mathrm{2}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{2}\right)} \\ $$$$=\frac{\left({n}+\mathrm{2}\right)\left(\mathrm{2}{n}+\mathrm{5}\right)\left(\left(\mathrm{2}{n}+\mathrm{4}\right)!\right)^{\mathrm{2}} }{\mathrm{4}^{\mathrm{2}{n}+\mathrm{3}} .\mathrm{3}\left(\left({n}+\mathrm{2}\right)!\right)^{\mathrm{4}} \left({n}+\mathrm{3}\right)}=\frac{\left({n}+\mathrm{1}+\mathrm{1}\right)\left(\mathrm{2}\left({n}+\mathrm{1}\right)+\mathrm{3}\right)\left(\mathrm{2}\left({n}+\mathrm{1}+\mathrm{1}\right)!\right)^{\mathrm{2}} }{\mathrm{4}^{\mathrm{2}\left({n}+\mathrm{1}\right)+\mathrm{1}} .\mathrm{3}\left(\left({n}+\mathrm{1}+\mathrm{1}\right)!\right)^{\mathrm{4}} \left({n}+\mathrm{1}+\mathrm{2}\right)} \\ $$$$={f}\left({n}+\mathrm{1}\right) \\ $$$$\Rightarrow\forall{n}\in\mathbb{N}\:\:\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{16}^{{k}} \left({k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{2}\right)}.\begin{pmatrix}{\mathrm{2}{k}}\\{{k}}\end{pmatrix}^{\mathrm{2}} =\frac{\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)\left(\left(\mathrm{2}{n}+\mathrm{2}\right)!\right)^{\mathrm{2}} }{\mathrm{4}^{\mathrm{2}{n}+\mathrm{1}} .\mathrm{3}\left(\left({n}+\mathrm{1}\right)!\right)^{\mathrm{4}} \left({n}+\mathrm{2}\right)} \\ $$$$\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{16}^{{k}} \left({k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{2}\right)}\begin{pmatrix}{\mathrm{2}{k}}\\{{k}}\end{pmatrix}^{\mathrm{2}} =\underset{{n}\rightarrow\infty} {\mathrm{lim}}{f}\left({n}\right) \\ $$$${Striling}\:{formula}\Rightarrow \\ $$$${n}!\sim\sqrt{\mathrm{2}\pi{n}}\left(\frac{{n}}{{e}}\right)^{{n}} \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)\left(\left(\mathrm{2}{n}+\mathrm{2}\right)!\right)^{\mathrm{2}} }{\mathrm{4}^{\mathrm{2}{n}+\mathrm{1}} .\mathrm{3}\left(\left({n}+\mathrm{1}\right)!\right)^{\mathrm{4}} \left({n}+\mathrm{2}\right)}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)\left(\sqrt{\mathrm{2}\pi\left(\mathrm{2}{n}+\mathrm{2}\right)}.\frac{\left(\mathrm{2}{n}+\mathrm{2}\right)^{\mathrm{2}{n}+\mathrm{2}} }{{e}^{\mathrm{2}{n}+\mathrm{2}} }\right)^{\mathrm{2}} }{\mathrm{4}^{\mathrm{2}{n}+\mathrm{1}} \left({n}+\mathrm{2}\right).\mathrm{3}.\left(\sqrt{\mathrm{2}\pi\left({n}+\mathrm{1}\right)}.\left(\frac{{n}+\mathrm{1}}{{e}}\right)^{{n}+\mathrm{1}} \right)^{\mathrm{4}} } \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right).\mathrm{2}\pi\left(\mathrm{2}{n}+\mathrm{2}\right).\frac{\left(\mathrm{2}{n}+\mathrm{2}\right)^{\mathrm{4}{n}+\mathrm{4}} }{{e}^{\mathrm{4}{n}+\mathrm{4}} }}{\mathrm{4}^{\mathrm{2}{n}+\mathrm{1}} \left({n}+\mathrm{2}\right).\mathrm{3}\left(\mathrm{2}\pi\left({n}+\mathrm{1}\right)\right)^{\mathrm{2}} .\left(\frac{{n}+\mathrm{1}}{{e}}\right)^{\mathrm{4}{n}+\mathrm{4}} } \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)\left(\mathrm{2}{n}+\mathrm{2}\right).\mathrm{2}\pi}{\mathrm{4}^{\mathrm{2}{n}+\mathrm{1}} \left({n}+\mathrm{2}\right).\mathrm{3}.\mathrm{4}\pi^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }.\frac{\mathrm{2}^{\mathrm{4}{n}+\mathrm{4}} \left({n}+\mathrm{1}\right)^{\mathrm{4}{n}+\mathrm{4}} }{{e}^{\mathrm{4}{n}+\mathrm{4}} }.\frac{{e}^{\mathrm{4}{n}+\mathrm{4}} }{\left({n}+\mathrm{1}\right)^{\mathrm{4}{n}+\mathrm{4}} } \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\left(\mathrm{2}{n}+\mathrm{3}\right)\left(\mathrm{2}{n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right).\mathrm{4}^{\mathrm{2}{n}+\mathrm{2}} }{\mathrm{4}^{\mathrm{2}{n}+\mathrm{1}} .\mathrm{6}\pi\left({n}+\mathrm{1}\right)\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{4}}{.\mathrm{3}.\mathrm{2}\pi}.\frac{\mathrm{2}{n}+\mathrm{3}}{{n}+\mathrm{1}}.\frac{\mathrm{2}{n}+\mathrm{2}}{{n}+\mathrm{1}}.\frac{{n}+\mathrm{1}}{{n}+\mathrm{2}}=\frac{\mathrm{4}}{\mathrm{6}\pi}.\mathrm{2}.\mathrm{2}=\frac{\mathrm{16}}{\mathrm{6}\pi}=\frac{\mathrm{8}}{\mathrm{3}\pi} \\ $$$$\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{16}^{{n}} \left({n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}\right)}\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}^{\mathrm{2}} =\frac{\mathrm{8}}{\mathrm{3}\pi} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 02/Oct/20

very astonishing thank you sir...excellent..

$${very}\:{astonishing}\: \\ $$$${thank}\:{you}\:{sir}...{excellent}.. \\ $$

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