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Question Number 47960 by Meritguide1234 last updated on 17/Nov/18

Answered by ajfour last updated on 17/Nov/18

$$L=\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}\underset{i=1}{\sum ^n}\underset{j=1}{\sum ^n}\frac{i+j}{i^2+j^2}$$$$=\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}\underset{i=1}{\sum ^n}\left(\underset{n\to 0}{\lim }\frac{1}{n}\underset{j=1}{\sum ^n}\frac{\frac{i}{n}+\frac{j}{n}}{\frac{i^2}{n^2}+\frac{j^2}{n^2}}\right)$$$$Let\frac{i}{n}=a$$$$=\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}\underset{i=1}{\sum ^n}{\int }_0^1\frac{a+x}{a^2+x^2}dx$$$$=\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}\underset{i=1}{\sum ^n}[{\int }_0^1\frac{a}{a^2+x^2}dx+$$$$\frac{1}{2}{\int }_0^1\frac{2xdx}{a^2+x^2}]$$$$=\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}\underset{i=1}{\sum ^n}[{\tan }^{-1}\frac{x}{a}{\mid }_0^1$$$$+\frac{1}{2}\ln \mid a^2+x^2{\mid }_0^1]$$$$=\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}\underset{i=1}{\sum ^n}({\tan }^{-1}\frac{1}{a}+\frac{1}{2}\ln \mid 1+\frac{1}{a^2}\mid )$$$$={\int }_0^1[{\tan }^{-1}\frac{1}{x}+\frac{1}{2}\ln (1+\frac{1}{x^2})]dx$$$$={[x{\tan }^{-1}\frac{1}{x}+\frac{1}{2}\ln (1+\frac{1}{x^2})]}_0^1$$$$-{\int }_0^1\left(\frac{1}{1+\frac{1}{x^2}}\right)\left(\frac{-1}{x^2}\right)xdx$$$$-\frac{1}{2}{\int }_0^1\left(\frac{1}{1+\frac{1}{x^2}}\right)\left(\frac{-2}{x^3}\right)xdx$$$$=\frac{\pi }{4}+\frac{\ln 2}{2}+\frac{1}{2}{\int }_0^1\frac{2xdx}{x^2+1}+{\int }_0^1\frac{dx}{1+x^2}$$$$L=\frac{\pi }{4}+\frac{\ln 2}{2}+\frac{\ln 2}{2}+\frac{\pi }{4}.$$

Commented by Meritguide1234 last updated on 17/Nov/18

$$beautifulsolution$$

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