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Question Number 143326 by SOMEDAVONG last updated on 13/Jun/21

$$\mathrm{L}=\underset{\mathrm{n}\to +\propto }{\lim }\frac{(2^3-1)(3^3-1)(4^3-1)...({\mathrm{n}}^3-1)}{(2^3+1)(3^3+1)(4^3+1)...({\mathrm{n}}^3+1)}$$

Answered by gsk2684 last updated on 13/Jun/21

$$\frac{{\mathrm{n}}^3-1}{{\mathrm{n}}^3+1}=\left(\frac{\mathrm{n}-1}{\mathrm{n}+1}\right).\left(\frac{{\mathrm{n}}^2+\mathrm{n}+1}{{\mathrm{n}}^2-\mathrm{n}+1}\right)$$$$\mathrm{L}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }(\frac{1}{3}.\frac{2}{4}.\frac{3}{5}.\frac{4}{6}...\frac{\mathrm{n}-2}{\mathrm{n}-1}.\frac{\mathrm{n}-1}{\mathrm{n}+1})(\frac{7}{3}.\frac{13}{7}.\frac{21}{13}.\frac{31}{21}...\frac{{\mathrm{n}}^2+\mathrm{n}+1}{{\mathrm{n}}^2-\mathrm{n}+1})$$$$=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\left(\frac{1.2}{\mathrm{n}-1.\mathrm{n}+1}\right)\left(\frac{{\mathrm{n}}^2+\mathrm{n}+1}{3}\right)$$$$=\frac{2}{3}\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{{\mathrm{n}}^2+\mathrm{n}+1}{{\mathrm{n}}^2-1}=\frac{2}{3}$$

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