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

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

Answered by john_santu last updated on 27/Feb/21

$$since\frac{k^3-1}{k^3+1}=\frac{(k-1)(k^2+k+1)}{(k+1)(k^2-k+1)}=\frac{(k-1)({(k+1)}^2-(k+1)+1)}{(k+1)(k^2-k+1)}$$$$weget\underset{k=2}{\prod ^n}\frac{k^3-1}{k^3+1}=\frac{2}{3}.\frac{n^2+n+1}{n^2+n}\underset{n\to \mathrm{\infty }}{\to }\frac{2}{3}.$$$$$$

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