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Question Number 67534 by mathmax by abdo last updated on 28/Aug/19

find the value of Π_(n=2) ^∞ ((n^3 −1)/(n^3 +1)) and Π_(n=1) ^∞ (1+(1/n^2 ))

$${find}\:{the}\:{value}\:{of}\:\:\prod_{{n}=\mathrm{2}} ^{\infty} \:\frac{{n}^{\mathrm{3}} −\mathrm{1}}{{n}^{\mathrm{3}} \:+\mathrm{1}} \\ $$$${and}\:\prod_{{n}=\mathrm{1}} ^{\infty} \left(\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right) \\ $$

Commented by ~ À ® @ 237 ~ last updated on 29/Aug/19

A=Π_(n=2) ^∞ ((n^3 −1)/(n^3 +1))= Π_(n=2) ^∞ (((n−1)(n^2 +n+1))/((n+1)(n^2 −n+1))) n^2 +n+1=(n−j)(n−j^_ ) with j=((−1)/2)+i(((√3) )/2) n^2 −n+1=[(n−1)+1]^2 −(n−1)=(n−1)^2 +2(n−1)+1−(n−1)=p^2 +p+1 =(p−j)(p−j^_ ) with p=n−1 n^2 −n+1=(n−1−j)(n−1−j^_ ) A_p =Π_(n=2) ^p ((n−1)/(n+1)) .Π_(n=2) ^p (((j−n)(j^_ −n))/([j−(n−1)][j^_ −(n−1)])) =(((p−1)!)/(((p+1)!)/(2!))).(((j−p)(j^_ −p))/((j−1)(j^_ −1))) =(2/((2−2Re(j)))) .(((j−p)(j^_ −p))/p^2 ) A=lim_(n→∞) A_p =(1/(1−Re(j))) = (2/3) Knowing that sin(πz)=πzΠ_(n=1) ^∞ (1−(z^2 /n^2 )) let take z=ia with a≠0 sin(iπa)=(iπa)Π_(n=1) ^∞ (1+(a^2 /n^2 )) Π_(n=1) ^∞ (1+(a^2 /n^2 ))=((sin(iπa))/(iπa)) =(((e^(−πa) −e^(πa) )/(2i))/(iπa)) =((sh(πa))/(πa))

$${A}=\underset{{n}=\mathrm{2}} {\overset{\infty} {\prod}}\:\frac{{n}^{\mathrm{3}} −\mathrm{1}}{{n}^{\mathrm{3}} +\mathrm{1}}=\:\underset{{n}=\mathrm{2}} {\overset{\infty} {\prod}}\:\frac{\left({n}−\mathrm{1}\right)\left({n}^{\mathrm{2}} +{n}+\mathrm{1}\right)}{\left({n}+\mathrm{1}\right)\left({n}^{\mathrm{2}} −{n}+\mathrm{1}\right)}\:\: \\ $$$${n}^{\mathrm{2}} +{n}+\mathrm{1}=\left({n}−{j}\right)\left({n}−\overset{\_} {{j}}\right)\:\:\:\:{with}\:\:{j}=\frac{−\mathrm{1}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{3}}\:}{\mathrm{2}} \\ $$$${n}^{\mathrm{2}} −{n}+\mathrm{1}=\left[\left({n}−\mathrm{1}\right)+\mathrm{1}\right]^{\mathrm{2}} −\left({n}−\mathrm{1}\right)=\left({n}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}\left({n}−\mathrm{1}\right)+\mathrm{1}−\left({n}−\mathrm{1}\right)={p}^{\mathrm{2}} +{p}+\mathrm{1}\:=\left({p}−{j}\right)\left({p}−\overset{\_} {{j}}\right)\:\:{with}\:\:{p}={n}−\mathrm{1} \\ $$$${n}^{\mathrm{2}} −{n}+\mathrm{1}=\left({n}−\mathrm{1}−{j}\right)\left({n}−\mathrm{1}−\overset{\_} {{j}}\right) \\ $$$${A}_{{p}} =\underset{{n}=\mathrm{2}} {\overset{{p}} {\prod}}\frac{{n}−\mathrm{1}}{{n}+\mathrm{1}}\:.\underset{{n}=\mathrm{2}} {\overset{{p}} {\prod}}\:\frac{\left({j}−{n}\right)\left(\overset{\_} {{j}}−{n}\right)}{\left[{j}−\left({n}−\mathrm{1}\right)\right]\left[\overset{\_} {{j}}−\left({n}−\mathrm{1}\right)\right]}\:=\frac{\left({p}−\mathrm{1}\right)!}{\frac{\left({p}+\mathrm{1}\right)!}{\mathrm{2}!}}.\frac{\left({j}−{p}\right)\left(\overset{\_} {{j}}−{p}\right)}{\left({j}−\mathrm{1}\right)\left(\overset{\_} {{j}}−\mathrm{1}\right)}\:=\frac{\mathrm{2}}{\left(\mathrm{2}−\mathrm{2}{Re}\left({j}\right)\right)}\:.\frac{\left({j}−{p}\right)\left(\overset{\_} {{j}}−{p}\right)}{{p}^{\mathrm{2}} }\: \\ $$$${A}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{A}_{{p}} \:=\frac{\mathrm{1}}{\mathrm{1}−{Re}\left({j}\right)}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\: \\ $$$${Knowing}\:\:{that}\:\:{sin}\left(\pi{z}\right)=\pi{z}\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{{z}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right)\:\:\:\:{let}\:\:{take}\:\:{z}={ia}\:\:{with}\:{a}\neq\mathrm{0}\: \\ $$$${sin}\left({i}\pi{a}\right)=\left({i}\pi{a}\right)\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}+\frac{{a}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right)\: \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}+\frac{{a}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right)=\frac{{sin}\left({i}\pi{a}\right)}{{i}\pi{a}}\:=\frac{\frac{{e}^{−\pi{a}} −{e}^{\pi{a}} }{\mathrm{2}{i}}}{{i}\pi{a}}\:=\frac{{sh}\left(\pi{a}\right)}{\pi{a}}\: \\ $$$$ \\ $$

Commented by Abdo msup. last updated on 29/Aug/19

thank you sir. let A =Π_(n=2) ^∞ ((n^3 −1)/(n^3 +1)) and A_n =Π_(k=2) ^n ((k^3 −1)/(k^(3 ) +1)) ⇒ A_n =Π_(k=2) ^n (((k−1)(k^2 +k+1))/((k+1)(k^2 −x+1))) =Π_(k=2) ^n ((k−1)/(k+1))∐_(k=2) ^n ((k^2 +k+1)/(k^2 −k +1)) ⇒ln(A_n )=Σ_(k=2) ^n ln(((k−1)/(k+1)))+Σ_(k=2) ^n ln(((k^2 +k+1)/(k^2 −k+1))) =Σ_(k=2) ^n {ln(k−1)−ln(k+1)) +Σ_(k=2) ^n { ln(k^2 +k+1)−ln(k^2 −k+1)} =Σ_(k=2) ^n {ln(k−1)−ln(k)}+Σ_(k=2) ^n {ln(k)−ln(k+1)} +Σ_(k=2) ^n (u_k −u_(k−1) ) with u_k =ln(k^2 +k+1) =ln(1)−ln(2)+ln(2)−ln(3)+....+ln(n−1)−ln(n) +ln(2)−ln(3)+ln(3)−ln(4)+...+ln(n)−ln(n+1) +u_n −u_1 =−ln(n)+ln(2)−ln(n+1) +ln(n^2 +n+1)−ln(3) =ln((2/3))+ln(((n^2 +n+1)/(n^2 +n)))→ln((2/3)) ⇒lim_(n→+∞) A_n =(2/3)

$${thank}\:{you}\:{sir}. \\ $$$${let}\:\:{A}\:=\prod_{{n}=\mathrm{2}} ^{\infty} \:\frac{{n}^{\mathrm{3}} −\mathrm{1}}{{n}^{\mathrm{3}} \:+\mathrm{1}}\:{and}\:{A}_{{n}} =\prod_{{k}=\mathrm{2}} ^{{n}} \:\frac{{k}^{\mathrm{3}} −\mathrm{1}}{{k}^{\mathrm{3}\:} +\mathrm{1}}\:\Rightarrow \\ $$$${A}_{{n}} \:\:=\prod_{{k}=\mathrm{2}} ^{{n}} \:\:\frac{\left({k}−\mathrm{1}\right)\left({k}^{\mathrm{2}} +{k}+\mathrm{1}\right)}{\left({k}+\mathrm{1}\right)\left({k}^{\mathrm{2}} −{x}+\mathrm{1}\right)} \\ $$$$=\prod_{{k}=\mathrm{2}} ^{{n}} \:\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}}\coprod_{{k}=\mathrm{2}} ^{{n}} \:\frac{{k}^{\mathrm{2}} \:+{k}+\mathrm{1}}{{k}^{\mathrm{2}} −{k}\:+\mathrm{1}} \\ $$$$\Rightarrow{ln}\left({A}_{{n}} \right)=\sum_{{k}=\mathrm{2}} ^{{n}} {ln}\left(\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}}\right)+\sum_{{k}=\mathrm{2}} ^{{n}} {ln}\left(\frac{{k}^{\mathrm{2}} \:+{k}+\mathrm{1}}{{k}^{\mathrm{2}} −{k}+\mathrm{1}}\right) \\ $$$$=\sum_{{k}=\mathrm{2}} ^{{n}} \:\left\{{ln}\left({k}−\mathrm{1}\right)−{ln}\left({k}+\mathrm{1}\right)\right) \\ $$$$+\sum_{{k}=\mathrm{2}} ^{{n}} \left\{\:{ln}\left({k}^{\mathrm{2}} \:+{k}+\mathrm{1}\right)−{ln}\left({k}^{\mathrm{2}} −{k}+\mathrm{1}\right)\right\} \\ $$$$=\sum_{{k}=\mathrm{2}} ^{{n}} \left\{{ln}\left({k}−\mathrm{1}\right)−{ln}\left({k}\right)\right\}+\sum_{{k}=\mathrm{2}} ^{{n}} \left\{{ln}\left({k}\right)−{ln}\left({k}+\mathrm{1}\right)\right\} \\ $$$$+\sum_{{k}=\mathrm{2}} ^{{n}} \left({u}_{{k}} −{u}_{{k}−\mathrm{1}} \right)\:\:{with}\:{u}_{{k}} ={ln}\left({k}^{\mathrm{2}} \:+{k}+\mathrm{1}\right) \\ $$$$={ln}\left(\mathrm{1}\right)−{ln}\left(\mathrm{2}\right)+{ln}\left(\mathrm{2}\right)−{ln}\left(\mathrm{3}\right)+....+{ln}\left({n}−\mathrm{1}\right)−{ln}\left({n}\right) \\ $$$$+{ln}\left(\mathrm{2}\right)−{ln}\left(\mathrm{3}\right)+{ln}\left(\mathrm{3}\right)−{ln}\left(\mathrm{4}\right)+...+{ln}\left({n}\right)−{ln}\left({n}+\mathrm{1}\right) \\ $$$$+{u}_{{n}} −{u}_{\mathrm{1}} \\ $$$$=−{ln}\left({n}\right)+{ln}\left(\mathrm{2}\right)−{ln}\left({n}+\mathrm{1}\right)\:+{ln}\left({n}^{\mathrm{2}} +{n}+\mathrm{1}\right)−{ln}\left(\mathrm{3}\right) \\ $$$$={ln}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)+{ln}\left(\frac{{n}^{\mathrm{2}} +{n}+\mathrm{1}}{{n}^{\mathrm{2}} \:+{n}}\right)\rightarrow{ln}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\:\Rightarrow{lim}_{{n}\rightarrow+\infty} {A}_{{n}} =\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$ \\ $$

Commented by Abdo msup. last updated on 29/Aug/19

we have sin(πz)=πz Π_(n=1) ^∞ (1−(z^2 /n^2 )) let z=ix (x real) ⇒sin(iπx) =iπxΠ_(n=1) ^∞ (1+(x^2 /n^2 )) ⇒ Π_(n=1) ^∞ (1+(x^2 /n^2 )) =((sin(iπx))/(iπx)) sinz =((e^(iz) −e^(−iz) )/(2i)) ⇒sin(iπx) =((e^(−πx) −e^(πx) )/(2i)) =−(1/i)sh(πx) ⇒Π_(n=1) ^∞ (1+(x^2 /n^2 )) =−(1/i) ((sh(πx))/(iπx)) =((sh(πx))/(πx)) x=1 ⇒ Π_(n=1) ^∞ (1+(1/n^2 )) =((sh(π))/π) .

$${we}\:{have}\:\:{sin}\left(\pi{z}\right)=\pi{z}\:\prod_{{n}=\mathrm{1}} ^{\infty} \left(\mathrm{1}−\frac{{z}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right) \\ $$$${let}\:{z}={ix}\:\:\:\:\left({x}\:{real}\right)\:\Rightarrow{sin}\left({i}\pi{x}\right)\:={i}\pi{x}\prod_{{n}=\mathrm{1}} ^{\infty} \left(\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right)\:\Rightarrow \\ $$$$\prod_{{n}=\mathrm{1}} ^{\infty} \left(\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right)\:=\frac{{sin}\left({i}\pi{x}\right)}{{i}\pi{x}}\: \\ $$$${sinz}\:=\frac{{e}^{{iz}} −{e}^{−{iz}} }{\mathrm{2}{i}}\:\Rightarrow{sin}\left({i}\pi{x}\right)\:=\frac{{e}^{−\pi{x}} −{e}^{\pi{x}} }{\mathrm{2}{i}} \\ $$$$=−\frac{\mathrm{1}}{{i}}{sh}\left(\pi{x}\right)\:\Rightarrow\prod_{{n}=\mathrm{1}} ^{\infty} \left(\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right)\:=−\frac{\mathrm{1}}{{i}}\:\frac{{sh}\left(\pi{x}\right)}{{i}\pi{x}} \\ $$$$=\frac{{sh}\left(\pi{x}\right)}{\pi{x}} \\ $$$${x}=\mathrm{1}\:\Rightarrow\:\prod_{{n}=\mathrm{1}} ^{\infty} \left(\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)\:=\frac{{sh}\left(\pi\right)}{\pi}\:. \\ $$

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