Menu Close

prove-n-1-2n-1-3-3-2n-1-2-2n-1-3-pi-6-

Tinku Tara 0 Comments
Pin




Question Number 219724 by Nicholas666 last updated on 01/May/25
prove; Π_(n=1) ^∞ (((2n+1)^3 −3(2n+1)+2)/((2n+1)^3 )) = (π/6)
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{prove}; \\ $$$$\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{3}\left(\mathrm{2}{n}+\mathrm{1}\right)+\mathrm{2}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} }\:=\:\frac{\pi}{\mathrm{6}} \\ $$$$ \\ $$
Answered by nothing48 last updated on 01/May/25
let (2n+1) be x, Simplify the numerator Π_(n=1) ^∞ (((2n+1)^3 −3(2n+1)+2)/((2n+1)^3 ))=Π_(n=1) ^∞ ((4n^2 (2n+3))/((2n+1)^3 )) Partial Product(P_N ) is Π_(n=1) ^∞ ((4n^2 (2n+3))/((2n+1)^3 ))=(Π_(n=1) ^N ((2n)/(2n+1)))^2 ×Π_(n=1) ^N ((2n+3)/(2n+1)) The second product is a telescoping Series Π_(n=1) ^N ((2n+3)/(2n+1))=(5/3)×(7/5)×(9/7)...((2N+3)/(2N+1))=((2N+3)/3) Now, Lets consider the first product and its relation to wallis product: (π/2)=lim_(N→∞) ((2^2 ×4^2 ...(2N)^2 )/(1^2 ×3^2 ...(2N−1)^2 ×(2N+1)))=lim_(N→∞) (((2×4...2N)/(1×3...(2N−1))))^2 (1/(2N+1)) Now, Π_(n=1) ^N ((2n)/(2n+1))=((2×4...2N)/(3×5...(2N−1)))∼(√(πN)) as N→∞ Therefore ((2×4...2N)/(3×5..(2N+1)))∼(√(π/(4N))) Substitute back this in P_N lim_(N→∞) P_N =lim_(N→∞) ((√(π/(4N))))^2 ×((2N+3)/3) lim_(N→∞) P_N =lim_(N→∞) (π/(4N))×((2N+3)/3)=(π/6) Hence Proved
$${let}\:\left(\mathrm{2}{n}+\mathrm{1}\right)\:{be}\:{x}, \\ $$$${Simplify}\:{the}\:{numerator} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{3}\left(\mathrm{2}{n}+\mathrm{1}\right)+\mathrm{2}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\frac{\mathrm{4}{n}^{\mathrm{2}} \left(\mathrm{2}{n}+\mathrm{3}\right)}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${Partial}\:{Product}\left({P}_{{N}} \right)\:{is} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\frac{\mathrm{4}{n}^{\mathrm{2}} \left(\mathrm{2}{n}+\mathrm{3}\right)}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} }=\left(\underset{{n}=\mathrm{1}} {\overset{{N}} {\prod}}\frac{\mathrm{2}{n}}{\mathrm{2}{n}+\mathrm{1}}\right)^{\mathrm{2}} ×\underset{{n}=\mathrm{1}} {\overset{{N}} {\prod}}\frac{\mathrm{2}{n}+\mathrm{3}}{\mathrm{2}{n}+\mathrm{1}} \\ $$$${The}\:{second}\:{product}\:{is}\:{a}\:{telescoping}\:{Series} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{{N}} {\prod}}\frac{\mathrm{2}{n}+\mathrm{3}}{\mathrm{2}{n}+\mathrm{1}}=\frac{\mathrm{5}}{\mathrm{3}}×\frac{\mathrm{7}}{\mathrm{5}}×\frac{\mathrm{9}}{\mathrm{7}}…\frac{\mathrm{2}{N}+\mathrm{3}}{\mathrm{2}{N}+\mathrm{1}}=\frac{\mathrm{2}{N}+\mathrm{3}}{\mathrm{3}} \\ $$$${Now}, \\ $$$${Lets}\:{consider}\:{the}\:{first}\:{product}\:{and}\:{its}\:{relation} \\ $$$${to}\:{wallis}\:{product}: \\ $$$$\frac{\pi}{\mathrm{2}}=\underset{{N}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{2}^{\mathrm{2}} ×\mathrm{4}^{\mathrm{2}} …\left(\mathrm{2}{N}\right)^{\mathrm{2}} }{\mathrm{1}^{\mathrm{2}} ×\mathrm{3}^{\mathrm{2}} …\left(\mathrm{2}{N}−\mathrm{1}\right)^{\mathrm{2}} ×\left(\mathrm{2}{N}+\mathrm{1}\right)}=\underset{{N}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{2}×\mathrm{4}…\mathrm{2}{N}}{\mathrm{1}×\mathrm{3}…\left(\mathrm{2}{N}−\mathrm{1}\right)}\right)^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}{N}+\mathrm{1}} \\ $$$${Now}, \\ $$$$\underset{{n}=\mathrm{1}} {\overset{{N}} {\prod}}\frac{\mathrm{2}{n}}{\mathrm{2}{n}+\mathrm{1}}=\frac{\mathrm{2}×\mathrm{4}…\mathrm{2}{N}}{\mathrm{3}×\mathrm{5}…\left(\mathrm{2}{N}−\mathrm{1}\right)}\sim\sqrt{\pi{N}}\:{as}\:{N}\rightarrow\infty \\ $$$${Therefore} \\ $$$$\frac{\mathrm{2}×\mathrm{4}…\mathrm{2}{N}}{\mathrm{3}×\mathrm{5}..\left(\mathrm{2}{N}+\mathrm{1}\right)}\sim\sqrt{\frac{\pi}{\mathrm{4}{N}}} \\ $$$${Substitute}\:{back}\:{this}\:{in}\:{P}_{{N}} \\ $$$$\underset{{N}\rightarrow\infty} {\mathrm{lim}}{P}_{{N}} =\underset{{N}\rightarrow\infty} {\mathrm{lim}}\left(\sqrt{\frac{\pi}{\mathrm{4}{N}}}\right)^{\mathrm{2}} ×\frac{\mathrm{2}{N}+\mathrm{3}}{\mathrm{3}} \\ $$$$\underset{{N}\rightarrow\infty} {\mathrm{lim}}{P}_{{N}} =\underset{{N}\rightarrow\infty} {\mathrm{lim}}\frac{\pi}{\mathrm{4}{N}}×\frac{\mathrm{2}{N}+\mathrm{3}}{\mathrm{3}}=\frac{\pi}{\mathrm{6}} \\ $$$${Hence}\:{Proved} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *