Question Number 212646 by MrGaster last updated on 20/Oct/24
$$ \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\underset{{j}={i}} {\overset{{i}} {\sum}}\frac{{i}\left({i}+{j}\right)}{\left({n}^{\mathrm{2}} +{i}^{\mathrm{2}} \right)\left({n}^{\mathrm{2}} +{j}^{\mathrm{2}} \right)} \\ $$
Commented by mr W last updated on 21/Oct/24
$$\underset{{j}=\mathrm{1}} {\overset{{i}} {\sum}}….. \\ $$
Answered by MrGaster last updated on 31/Oct/24
![lim_(n→∞) Σ_(i=1) ^n ((2i^2 )/((n^2 +i^2 )^2 )) lim_(n→∞) (1/n)Σ_(i=1) ^n ((2((i/n))^2 )/((1+((i/n))^2 )))∙(1/n) ∫_0 ^1 ((2x^2 )/((1+x^2 )^2 ))dx [−(x/(1+x^2 ))+arctan(x)]_0 ^1 −(1/2)+(π/4)](https://www.tinkutara.com/question/Q213140.png)
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{2}{i}^{\mathrm{2}} }{\left({n}^{\mathrm{2}} +{i}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}}\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{2}\left(\frac{{i}}{{n}}\right)^{\mathrm{2}} }{\left(\mathrm{1}+\left(\frac{{i}}{{n}}\right)^{\mathrm{2}} \right)}\centerdot\frac{\mathrm{1}}{{n}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$$$\left[−\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }+\mathrm{arctan}\left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}} \\ $$