Question Number 144067 by SOMEDAVONG last updated on 21/Jun/21
$$\mathrm{L}=\underset{\mathrm{n}\rightarrow+\propto} {\mathrm{lim}}\left(\frac{\mathrm{1}^{\mathrm{2}} }{\mathrm{n}^{\mathrm{3}} +\mathrm{1}^{\mathrm{3}} }\:+\:\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{n}^{\mathrm{3}} +\mathrm{2}^{\mathrm{3}} }\:+\:…+\:\frac{\mathrm{n}^{\mathrm{3}} }{\mathrm{n}^{\mathrm{3}} +\mathrm{n}^{\mathrm{3}} }\right)=? \\ $$
Answered by Dwaipayan Shikari last updated on 21/Jun/21
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{k}^{\mathrm{2}} }{{n}^{\mathrm{3}} +{k}^{\mathrm{3}} }=\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\left(\frac{{k}}{{n}}\right)^{\mathrm{2}} }{\mathrm{1}+\left(\frac{{k}}{{n}}\right)^{\mathrm{3}} } \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{3}} }{dx}=\frac{\mathrm{1}}{\mathrm{3}}{log}\left(\mathrm{2}\right) \\ $$$$ \\ $$