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Question Number 27664 by abdo imad last updated on 12/Jan/18

$$letgivethesequence{V}_n=\prod _{k=1}^{k=n}{(1+\frac{k^2}{n^2})}^{\frac{1}{n}}$$ $$findthevalueoflim{}_{n->\propto }{V}_n.$$

Commented byabdo imad last updated on 14/Jan/18

$$wehaveln\left(V_n\right)=\frac{1}{n}ln\left(\prod _{k=1}^n(1+\frac{k^2}{n^2})\right)$$ $$=\frac{1}{n}\sum _{k=1}^{n}ln(1+\frac{k^2}{n^2})and{lim}_{n->\propto }ln\left(V_n\right)l={\int }_0^{1}ln(1+x^2)dx$$ $$for/t/<1{ln}^{,}(1+t)=\sum _{n=0}^{\propto }{(-1)}^n{t}^n$$ $$\Rightarrow ln(1+t)=\sum _{n=0}^{\propto }\frac{{(-1)}^n{t}^{n+1}}{n+1}=\sum _{n=1}^{\propto }{(-1)}^{n-1}\frac{t^n}{n}$$ $$\Rightarrow ln(1+x^2)=\sum _{n=1}^{\propto }{(-1)}^{n-1}\frac{x^{2n}}{n}and$$ $${\int }_0^{1}ln(1+x^2)dx=\sum _{n=1}^{\propto }\frac{{(-1)}^{n-1}}{n}{\int }_0^1{x}^{2n}dx$$ $$=\sum _{n=1}^{\propto }\frac{{(-1)}^{n-1}}{n(2n+1)}$$ $$\frac{1}{2}{\int }_0^{1}ln(1+x^2)dx=\sum _{n=1}^{\propto }(\frac{1}{2n}-\frac{1}{2n+1}){(-1)}^{n-1}$$ $$=\frac{1}{2}\sum _{n=1}^{\propto }\frac{{(-1)}^{n-1}}{n}+\sum _{n=1}^{\propto }\frac{{(-1)}^n}{2n+1}but$$ $$\sum _{n=1}^{\propto }\frac{{(-1)}^{n-1}}{n}=ln2$$ $$\sum _{n=1}^{\propto }\frac{{(-1)}^n}{2n+1}=\frac{\pi }{4}-1$$ $${\int }_0^{1}ln(1+x^2)dx=ln2+\frac{\pi }{2}-2$$ $${lim}_{n->\propto }ln\left(V_n\right)=ln2+\frac{\pi }{2}-2$$ $$\Rightarrow {lim}_{n->{\propto }^{}}{V}_n=2e^{\frac{\pi }{2}-2}.$$

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