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Q-158528-P-n-1-n-1-3-1-n-1-3-1-P-n-1-n-1-3-1-3-n-1-3-1-3-P-n-1-n-1-1-n-2-2n-1-n-1-1-n-1-1-n-2-2n-1-n-1-1-P-n-1-




Question Number 158751 by puissant last updated on 08/Nov/21
Q 158528          P=Π_(n=1) ^∞ ((((n+1)^3 −1)/((n+1)^3 +1)))  ⇒ P = Π_(n=1) ^∞ ((((n+1)^3 −1^3 )/((n+1)^3 +1^3 )))  ⇒ P = Π_(n=1) ^∞ {(((n+1−1)(n^2 +2n+1+n+1+1))/((n+1+1)(n^2 +2n+1−n−1+1)))}  ⇒ P = Π_(n=1) ^∞ {(n/(n+2))}•Π_(n=1) ^∞ {((n^2 +3n+3)/(n^2 +n+1))}  = lim_(n→∞) Π_(k=1) ^n {(k/(k+2))}•lim_(n→∞) Π_(k=1) ^n {((k^2 +3k+3)/(k^2 +k+1))}  = lim_(n→∞) {(1/3)•(2/4)•(3/5)•...•(n/(n+2))}×lim_(n→∞) {(7/3)•((13)/7)•...•((n^2 +3n+3)/(n^2 +n+1))}  =2lim_(n→∞) {(1/((n+1)(n+2)))}×(1/3)lim_(n→∞) {n^2 +3n+3}  =(2/3)lim_(n→∞) {((n^2 +3n+3)/(n^2 +3n+2))} = (2/3)lim_(n→∞) {((1+(3/n)+(3/n^2 ))/(1+(3/n)+(2/n^2 )))} = (2/3).                             P = Π_(n=1) ^∞ ((((n+1)^3 −1)/((n+1)^3 +1))) = (2/3)..                                  ...............Le puissant...............
Q158528P=n=1((n+1)31(n+1)3+1)P=n=1((n+1)313(n+1)3+13)P=n=1{(n+11)(n2+2n+1+n+1+1)(n+1+1)(n2+2n+1n1+1)}P=n=1{nn+2}n=1{n2+3n+3n2+n+1}=limnnk=1{kk+2}limnnk=1{k2+3k+3k2+k+1}=limn{132435nn+2}×limn{73137n2+3n+3n2+n+1}=2limn{1(n+1)(n+2)}×13limn{n2+3n+3}=23limn{n2+3n+3n2+3n+2}=23limn{1+3n+3n21+3n+2n2}=23.P=n=1((n+1)31(n+1)3+1)=23..Lepuissant

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