Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 66172 by mathmax by abdo last updated on 10/Aug/19

$$$$$$let{A}_n=\prod _{k=1}^n(1+\frac{k^2}{n^2})calculate{lim}_{n\to +\mathrm{\infty }}\frac{ln\left(A_n\right)}{n}$$

Commented by Prithwish sen last updated on 11/Aug/19

$${\lim }_{\mathrm{n}\to \mathrm{\infty }}\frac{1}{\mathrm{n}}\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\ln [1+{\left(\frac{\mathrm{k}}{\mathrm{n}}\right)}^2]={\int }_0^1\ln (1+{\mathrm{x}}^2)\mathrm{dx}$$$$={[\mathrm{xln}(1+{\mathrm{x}}^2)-2\mathrm{x}+{2\tan }^{-1}\mathrm{x}]}_0^1=\ln 2-2+\frac{\pi }{2}$$$$\mathrm{please}\mathrm{check}.$$

Commented by mathmax by abdo last updated on 10/Aug/19

$$wehaveln\left(A_n\right)=\sum _{k=1}^{n}ln(1+\frac{k^2}{n^2})\Rightarrow \frac{ln\left(A_n\right)}{n}=\frac{1}{n}\sum _{k=1}^{n}ln(1+{\left(\frac{k}{n}\right)}^2)$$$$wegetaRiemansum\Rightarrow {lim}_{n\to +\mathrm{\infty }}\frac{ln\left(A_n\right)}{n}={\int }_0^{1}ln(1+x^2)dx$$$$byparts{\int }_0^{1}ln(1+x^2)dx={\left[xln(1+x^2)\right]}_0^1-{\int }_0^{1}x\frac{2x}{1+x^2}dx$$$$=ln\left(2\right)-2{\int }_0^1\frac{x^2+1-1}{x^2+1}dx=ln\left(2\right)-2+2{\int }_0^1\frac{dx}{1+x^2}$$$$=ln\left(2\right)-2+2{\left[arctanx\right]}_0^1=ln\left(2\right)-2+2\times \frac{\pi }{4}=ln\left(2\right)-2+\frac{\pi }{2}$$$$finally{lim}_{n\to +\mathrm{\infty }}\frac{ln\left(A_n\right)}{n}=\frac{\pi }{2}+ln\left(2\right)-2.$$

Commented by Prithwish sen last updated on 10/Aug/19

$$\mathrm{thanks}\mathrm{sir}.$$

Commented by mathmax by abdo last updated on 10/Aug/19

$$youarewelcomesir.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com