L-lim-n-2-3-1-3-3-1-4-3-1-n-3-1-2-3-1-3-3-1-4-3-1-n-3-1-

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

L = \lim_{n \to + \propto} \frac{(2^{3} - 1) (3^{3} - 1) (4^{3} - 1) \dots (n^{3} - 1)}{(2^{3} + 1) (3^{3} + 1) (4^{3} + 1) \dots (n^{3} + 1)}

Answered by gsk2684 last updated on 13/Jun/21

\frac{n^{3} - 1}{n^{3} + 1} = (\frac{n - 1}{n + 1}) . (\frac{n^{2} + n + 1}{n^{2} - n + 1})

L = \lim_{n \to \infty} (\frac{1}{3} . \frac{2}{4} . \frac{3}{5} . \frac{4}{6} \dots \frac{n - 2}{n - 1} . \frac{n - 1}{n + 1}) (\frac{7}{3} . \frac{13}{7} . \frac{21}{13} . \frac{31}{21} \dots \frac{n^{2} + n + 1}{n^{2} - n + 1})

= \lim_{n \to \infty} (\frac{1.2}{n - 1 . n + 1}) (\frac{n^{2} + n + 1}{3})

= \frac{2}{3} \lim_{n \to \infty} \frac{n^{2} + n + 1}{n^{2} - 1} = \frac{2}{3}

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