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Question Number 108320 by Khalmohmmad last updated on 16/Aug/20

Answered by Dwaipayan Shikari last updated on 16/Aug/20

$$T_n=n.2n.4n=8n^3$$$$And$$$$T_n^{{}^{\prime }}=n.3n.9n=27n^3$$$${\left(\frac{1.2.4+2.4.8+...}{1.3.9+2.6.8+...}\right)}^{\frac{1}{3}}=\underset{n\to \mathrm{\infty }}{\lim }{\left(\frac{\sum ^{n}8n^3}{\sum ^{n}27n^3}\right)}^{\frac{1}{3}}=\underset{n\to \mathrm{\infty }}{\lim }{(\frac{8}{27}.\frac{{\left(\frac{n(n+1)}{2}\right)}^2}{{\left(\frac{n(n+1)}{2}\right)}^2})}^{\frac{1}{3}}=\frac{2}{3}$$

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