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Question Number 172189 by Mikenice last updated on 23/Jun/22

Answered by Mathspace last updated on 24/Jun/22

$$let{u}_n=\prod _{k=1}^n{\left(\phi (a+\frac{kb}{n})\right)}^{\frac{1}{n}}$$$$\Rightarrow {lim}_{n\to +\mathrm{\infty }}ln\left(u_n\right)$$$$={lim}_{n\to +\mathrm{\infty }}\frac{1}{n}ln(\phi (a+\frac{kb}{n})$$$$=\frac{1}{b}{lim}_{n\to +\mathrm{\infty }}\frac{a+b-a}{n}ln\left(\phi (a+\frac{k(a+b-a)}{n})\right)$$$$=\frac{1}{b}{\int }_a^{a+b-a}ln\left(\phi \left(x\right)\right)dx=\lambda \Rightarrow$$$${lim}_{n\to +\mathrm{\infty }}{u}_n=e^{\lambda }$$

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