Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 182256 by mathlove last updated on 06/Dec/22

Answered by Ar Brandon last updated on 06/Dec/22

$$\mathcal{L}=\underset{n\to \mathrm{\infty }}{\lim }(\underset{m\to \mathrm{\infty }}{\lim }\left(\underset{r=1}{\sum ^n}\left(\underset{k=1}{\sum ^{mr}}\frac{{mn}^2}{(m^2{n}^2+k^2)(n^2+r^2)}\right)\right)$$$$=\underset{n\to \mathrm{\infty }}{\lim }\left(\underset{r=1}{\sum ^n}\underset{m\to \mathrm{\infty }}{\lim }\left(\frac{1}{m}\underset{k=1}{\sum ^{mr}}\frac{n^2}{(n^2+\frac{k^2}{m^2})(n^2+r^2)}\right)\right)$$$$=\underset{n\to \mathrm{\infty }}{\lim }\left(\underset{r=1}{\sum ^n}\frac{n^2}{n^2+r^2}{\int }_0^r\frac{dx}{(n^2+x^2)}\right)=\underset{n\to \mathrm{\infty }}{\lim }\left(\underset{r=1}{\sum ^n}\frac{n^2}{n^2+r^2}{\left[\frac{1}{n}\arctan \left(\frac{x}{n}\right)\right]}_0^r\right)$$$$=\underset{n\to \mathrm{\infty }}{\lim }\left(\underset{r=1}{\sum ^n}\frac{n}{n^2+r^2}\arctan \left(\frac{r}{n}\right)\right)=\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}\underset{r=1}{\sum ^n}\frac{1}{1+\frac{r^2}{n^2}}\cdot \arctan \left(\frac{r}{n}\right)$$$$={\int }_0^1\frac{\arctan x}{1+x^2}dx={\int }_0^1\arctan \left(x\right)d\left(\arctan x\right)={\left[\frac{{\left(\arctan x\right)}^2}{2}\right]}_0^1=\frac{{\pi }^2}{32}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com