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Question Number 103436 by bramlex last updated on 15/Jul/20

Π_(n = 3) ^∞ (1−(4/n^2 )) = ?

$$\underset{\mathrm{n}\:=\:\mathrm{3}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{\mathrm{4}}{\mathrm{n}^{\mathrm{2}} }\right)\:=\:? \\ $$

Answered by Worm_Tail last updated on 15/Jul/20

Π_3 ^(oo) (1−(4/n^2 ))=a (1−(4/3^2 ))(1−(4/4^2 ))(1+(4/5^2 ))...=a ln((1−(4/3^2 ))(1−(4/4^2 ))(1+(4/5^2 ))...)=ln(a) ln(1−(4/3^2 ))+ln(1−(4/4^2 ))+ln(1+(4/5^2 ))...=ln(a) Σ_(n=3) ^(oo) (ln(1−(4/n^2 )))=lna Σ_(n=3) ^(oo) (ln((((n+2)(n−2))/n^2 )))=lna Σ_(n=3) ^(oo) (ln(n+2)+ln(n−2)−2ln(n))=lna Σ_(n=3) ^(oo) ln(n+2)+Σ_(n=3) ^(oo) ln(n−2)−2Σ_(n=3) ^(oo) ln(n)=lna Σ_(n=3) ^(oo) ln(n+2)+(ln(1)+ln(2)+ln(3)+ln(4)+Σ_(n=7) ^(oo) ln(n−2))−2(ln(3)+ln(4)+Σ_(n=5) ^(oo) ln(n))=lna Σ_(n=3) ^(oo) ln(n+2)=Σ_(n=7) ^(oo) ln(n−2)=Σ_(n=5) ^(oo) ln(n)=s s+(ln(1)+ln(2)+ln(3)+ln(4)+s)−2(ln(3)+ln(4)+s)=lna s+ln(1)+ln(2)+ln(3)+ln(4)+s−2ln(3)−2ln(4)−2s=lna s+ln(1)+ln(2)+ln(3)+ln(4)+s−2ln(3)−2ln(4)−2s=lna s+s−2s+ln(1)+ln(2)−ln(3)−ln(4)=ln(a) ln((2/(12)))=ln(a) (2/(12))=a a=(1/6)

$$\underset{\mathrm{3}} {\overset{{oo}} {\prod}}\left(\mathrm{1}−\frac{\mathrm{4}}{{n}^{\mathrm{2}} }\right)={a}\: \\ $$$$\left(\mathrm{1}−\frac{\mathrm{4}}{\mathrm{3}^{\mathrm{2}} }\right)\left(\mathrm{1}−\frac{\mathrm{4}}{\mathrm{4}^{\mathrm{2}} }\right)\left(\mathrm{1}+\frac{\mathrm{4}}{\mathrm{5}^{\mathrm{2}} }\right)...={a} \\ $$$$\:\:{ln}\left(\left(\mathrm{1}−\frac{\mathrm{4}}{\mathrm{3}^{\mathrm{2}} }\right)\left(\mathrm{1}−\frac{\mathrm{4}}{\mathrm{4}^{\mathrm{2}} }\right)\left(\mathrm{1}+\frac{\mathrm{4}}{\mathrm{5}^{\mathrm{2}} }\right)...\right)={ln}\left({a}\right) \\ $$$$\:\:{ln}\left(\mathrm{1}−\frac{\mathrm{4}}{\mathrm{3}^{\mathrm{2}} }\right)+{ln}\left(\mathrm{1}−\frac{\mathrm{4}}{\mathrm{4}^{\mathrm{2}} }\right)+{ln}\left(\mathrm{1}+\frac{\mathrm{4}}{\mathrm{5}^{\mathrm{2}} }\right)...={ln}\left({a}\right) \\ $$$$\:\:\:\underset{{n}=\mathrm{3}} {\overset{{oo}} {\sum}}\left({ln}\left(\mathrm{1}−\frac{\mathrm{4}}{{n}^{\mathrm{2}} }\right)\right)={lna}\: \\ $$$$\:\:\:\underset{{n}=\mathrm{3}} {\overset{{oo}} {\sum}}\left({ln}\left(\frac{\left({n}+\mathrm{2}\right)\left({n}−\mathrm{2}\right)}{{n}^{\mathrm{2}} }\right)\right)={lna}\: \\ $$$$\:\:\:\underset{{n}=\mathrm{3}} {\overset{{oo}} {\sum}}\left({ln}\left({n}+\mathrm{2}\right)+{ln}\left({n}−\mathrm{2}\right)−\mathrm{2}{ln}\left({n}\right)\right)={lna}\: \\ $$$$\:\:\:\underset{{n}=\mathrm{3}} {\overset{{oo}} {\sum}}{ln}\left({n}+\mathrm{2}\right)+\underset{{n}=\mathrm{3}} {\overset{{oo}} {\sum}}{ln}\left({n}−\mathrm{2}\right)−\mathrm{2}\underset{{n}=\mathrm{3}} {\overset{{oo}} {\sum}}{ln}\left({n}\right)={lna}\: \\ $$$$\:\:\:\underset{{n}=\mathrm{3}} {\overset{{oo}} {\sum}}{ln}\left({n}+\mathrm{2}\right)+\left({ln}\left(\mathrm{1}\right)+{ln}\left(\mathrm{2}\right)+{ln}\left(\mathrm{3}\right)+{ln}\left(\mathrm{4}\right)+\underset{{n}=\mathrm{7}} {\overset{{oo}} {\sum}}{ln}\left({n}−\mathrm{2}\right)\right)−\mathrm{2}\left({ln}\left(\mathrm{3}\right)+{ln}\left(\mathrm{4}\right)+\underset{{n}=\mathrm{5}} {\overset{{oo}} {\sum}}{ln}\left({n}\right)\right)={lna}\: \\ $$$$\:\:\:\underset{{n}=\mathrm{3}} {\overset{{oo}} {\sum}}{ln}\left({n}+\mathrm{2}\right)=\underset{{n}=\mathrm{7}} {\overset{{oo}} {\sum}}{ln}\left({n}−\mathrm{2}\right)=\underset{{n}=\mathrm{5}} {\overset{{oo}} {\sum}}{ln}\left({n}\right)={s} \\ $$$$\:\:\:{s}+\left({ln}\left(\mathrm{1}\right)+{ln}\left(\mathrm{2}\right)+{ln}\left(\mathrm{3}\right)+{ln}\left(\mathrm{4}\right)+{s}\right)−\mathrm{2}\left({ln}\left(\mathrm{3}\right)+{ln}\left(\mathrm{4}\right)+{s}\right)={lna}\: \\ $$$$\:\:\:{s}+{ln}\left(\mathrm{1}\right)+{ln}\left(\mathrm{2}\right)+{ln}\left(\mathrm{3}\right)+{ln}\left(\mathrm{4}\right)+{s}−\mathrm{2}{ln}\left(\mathrm{3}\right)−\mathrm{2}{ln}\left(\mathrm{4}\right)−\mathrm{2}{s}={lna}\: \\ $$$$\:\:\:{s}+{ln}\left(\mathrm{1}\right)+{ln}\left(\mathrm{2}\right)+{ln}\left(\mathrm{3}\right)+{ln}\left(\mathrm{4}\right)+{s}−\mathrm{2}{ln}\left(\mathrm{3}\right)−\mathrm{2}{ln}\left(\mathrm{4}\right)−\mathrm{2}{s}={lna}\: \\ $$$${s}+{s}−\mathrm{2}{s}+{ln}\left(\mathrm{1}\right)+{ln}\left(\mathrm{2}\right)−{ln}\left(\mathrm{3}\right)−{ln}\left(\mathrm{4}\right)={ln}\left({a}\right) \\ $$$${ln}\left(\frac{\mathrm{2}}{\mathrm{12}}\right)={ln}\left({a}\right) \\ $$$$\frac{\mathrm{2}}{\mathrm{12}}={a} \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$

Commented by bobhans last updated on 15/Jul/20

waw..via logarithm

$${waw}..{via}\:{logarithm} \\ $$

Answered by bemath last updated on 15/Jul/20

Commented by bobhans last updated on 15/Jul/20

cooll graphic

$${cooll}\:{graphic}\: \\ $$

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