Calculate-lim-n-k-2-n-k-3-1-k-3-1-

Question Number 134067 by bramlexs22 last updated on 27/Feb/21

Calculate \lim_{n \to \infty} \underset{k = 2}{\prod^{n}} \frac{k^{3} - 1}{k^{3} + 1} = ?

Answered by john_santu last updated on 27/Feb/21

s i n c e \frac{k^{3} - 1}{k^{3} + 1} = \frac{(k - 1) (k^{2} + k + 1)}{(k + 1) (k^{2} - k + 1)} = \frac{(k - 1) ({(k + 1)}^{2} - (k + 1) + 1)}{(k + 1) (k^{2} - k + 1)}

w e g e t \underset{k = 2}{\prod^{n}} \frac{k^{3} - 1}{k^{3} + 1} = \frac{2}{3} . \frac{n^{2} + n + 1}{n^{2} + n} \underset{n \to \infty}{\to} \frac{2}{3} .