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Question Number 111754 by mathmax by abdo last updated on 04/Sep/20

$$$$$$\mathrm{calculate}{\lim }_{\mathrm{n}\to +\mathrm{\infty }}\sum _{\mathrm{k}=1}^{2\mathrm{n}}\frac{\mathrm{k}}{{\mathrm{k}}^2+{\mathrm{n}}^2}$$

Answered by Dwaipayan Shikari last updated on 04/Sep/20

$$\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}\sum ^{2n}\frac{k}{\frac{k^2}{n}+n}=\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}\underset{k=1}{\sum ^{2n}}\frac{1}{\frac{k}{n}+\frac{n}{k}}={\int }_0^2\frac{1}{x+\frac{1}{x}}dx$$$$=\frac{1}{2}{\int }_0^2\frac{2x}{x^2+1}dx$$$$={\left[\frac{1}{2}log(x^2+1)\right]}_0^2$$$$=\frac{1}{2}log\left(5\right)$$

Answered by mathmax by abdo last updated on 07/Sep/20

$${\mathrm{A}}_{\mathrm{n}}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{\mathrm{k}}{{\mathrm{k}}^2+{\mathrm{n}}^2}+\sum _{\mathrm{k}=\mathrm{n}+1}^{2\mathrm{n}}\frac{\mathrm{k}}{{\mathrm{k}}^2+{\mathrm{n}}^2}={\mathrm{S}}_1^{\mathrm{n}}+{\mathrm{S}}_2^{\mathrm{n}}$$$${\mathrm{S}}_1^{\mathrm{n}}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{\mathrm{k}}{{\mathrm{k}}^2(1+\frac{{\mathrm{n}}^2}{{\mathrm{k}}^2})}=\frac{1}{\mathrm{n}}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{\mathrm{n}}{\mathrm{k}(1+\frac{1}{\frac{{\mathrm{n}}^2}{{\mathrm{k}}^2}})}\to {\int }_0^1\frac{\mathrm{x}}{1+\frac{1}{\mathrm{x}}}\mathrm{dx}$$$$={\int }_0^1\frac{{\mathrm{x}}^2}{\mathrm{x}+1}\mathrm{dx}={\int }_0^1\frac{{\mathrm{x}}^2-1+1}{\mathrm{x}+1}\mathrm{dx}={\int }_0^1(\mathrm{x}-1)\mathrm{dx}+{\int }_0^1\frac{\mathrm{dx}}{\mathrm{x}+1}$$$$={[\frac{{\mathrm{x}}^2}{2}-\mathrm{x}]}_0^1+{\left[\ln (\mathrm{x}+1)\right]}_0^1=-\frac{1}{2}+\ln \left(2\right)$$$${\mathrm{S}}_2^{\mathrm{n}}=\sum _{\mathrm{k}=\mathrm{n}+1}^{2\mathrm{n}}\frac{\mathrm{k}}{{\mathrm{k}}^2+{\mathrm{n}}^2}{=}_{\mathrm{k}-\mathrm{n}=\mathrm{p}}\sum _{\mathrm{p}=1}^{\mathrm{n}}\frac{\mathrm{n}+\mathrm{p}}{{(\mathrm{n}+\mathrm{p})}^2+{\mathrm{n}}^2}$$$$=\sum _{\mathrm{p}=1}^{\mathrm{n}}\frac{\mathrm{n}+\mathrm{p}}{{\mathrm{n}}^2+2\mathrm{np}+{\mathrm{p}}^2+{\mathrm{n}}^2}=\sum _{\mathrm{p}=1}^{\mathrm{n}}\frac{\mathrm{n}+\mathrm{p}}{{2\mathrm{n}}^2+2\mathrm{np}+{\mathrm{p}}^2}$$$$=\sum _{\mathrm{p}=1}^{\mathrm{n}}\frac{\mathrm{n}(1+\frac{\mathrm{p}}{\mathrm{n}})}{{\mathrm{n}}^2(2+\frac{2\mathrm{p}}{\mathrm{n}}+\frac{{\mathrm{p}}^2}{{\mathrm{n}}^2})}=\frac{1}{\mathrm{n}}\sum _{\mathrm{p}=1}^{\mathrm{n}}\frac{1+\frac{\mathrm{p}}{\mathrm{n}}}{{\left(\frac{\mathrm{p}}{\mathrm{n}}\right)}^2+2\left(\frac{\mathrm{p}}{\mathrm{n}}\right)+2}\to {\int }_0^1\frac{1+\mathrm{x}}{{\mathrm{x}}^2+2\mathrm{x}+2}\mathrm{dx}$$$$=\frac{1}{2}{\int }_0^1\frac{2\mathrm{x}+2}{{\mathrm{x}}^2+2\mathrm{x}+2}\mathrm{dx}=\frac{1}{2}{\left[\ln ({\mathrm{x}}^2+2\mathrm{x}+2)\right]}_0^1=\frac{1}{2}(\ln \left(5\right)-\ln \left(2\right))\Rightarrow$$$${\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{A}}_{\mathrm{n}}=\ln \left(2\right)-\frac{1}{2}+\frac{1}{2}\ln 5-\frac{1}{2}\ln 2=\frac{1}{2}\ln 2+\frac{1}{2}\ln 5-\frac{1}{2}$$$$=\ln \left(\sqrt{10}\right)-\frac{1}{2}$$

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