Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 46840 by maxmathsup by imad last updated on 01/Nov/18

$$find{lim}_{n\to +\mathrm{\infty }}\sum _{k=1}^n\frac{n^2+k^2}{n^3+k^3}$$

Commented by maxmathsup by imad last updated on 03/Nov/18

$$let{S}_n=\sum _{k=1}^n\frac{n^2+k^2}{n^3+k^3}\Rightarrow S_n=\sum _{k=1}^n\frac{n^2(1+\frac{k^2}{n^2})}{n^3(1+\frac{k^3}{n^3})}=\sum _{k=1}^n\frac{1+{\left(\frac{k}{n}\right)}^2}{1+{\left(\frac{k}{n}\right)}^3}$$$$so{S}_{n}isaRiemansumand{lim}_{n\to +\mathrm{\infty }}{S}_n={\int }_0^1\frac{1+x^2}{1+x^3}dx$$$$={\int }_0^1\frac{dx}{1+x^3}+{\int }_0^1\frac{x^2}{1+x^3}dxbut{\int }_0^1\frac{x^2}{1+x^3}dx={\left[\frac{1}{3}ln(1+x^3)\right]}_0^1=\frac{ln\left(2\right)}{3}$$$$letdecomposeF\left(x\right)=\frac{1}{x^3+1}=\frac{1}{(x+1)(x^2-x+1)}=\frac{a}{x+1}+\frac{bx+c}{x^2-x+1}$$$$a={lim}_{x\to -1}(x+1)F\left(x\right)=\frac{1}{3}$$$${lim}_{x\to +\mathrm{\infty }}xF\left(x\right)=0=a+b\Rightarrow b=-\frac{1}{3}\Rightarrow F\left(x\right)=\frac{1}{3(x+1)}+\frac{-\frac{1}{3}x+c}{x^2-x+1}$$$$F\left(0\right)=1=\frac{1}{3}+c\Rightarrow c=\frac{2}{3}\Rightarrow F\left(x\right)=\frac{1}{3(x+1)}-\frac{1}{3}\frac{x-2}{x^2-x+1}\Rightarrow$$$${\int }_0^{1}F\left(x\right)=\frac{1}{3}{[ln\mid x+1\mid ]}_0^1-\frac{1}{6}{\int }_0^1\frac{2x-1-3}{x^2-x+1}dx$$$$=\frac{ln\left(2\right)}{3}-\frac{1}{6}{[ln\mid x^2-x+1\mid ]}_0^1+\frac{1}{2}{\int }_0^1\frac{dx}{x^2-x+1}$$$$=\frac{ln\left(2\right)}{3}+\frac{1}{2}{\int }_0^1\frac{dx}{{(x-\frac{1}{2})}^2+\frac{3}{4}}{=}_{x-\frac{1}{2}=\frac{\sqrt{3}}{2}t}\frac{ln\left(2\right)}{3}+\frac{1}{2}{\int }_{-\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}}\frac{4}{3}\frac{1}{t^2+1}\frac{\sqrt{3}}{2}dt$$$$=\frac{ln\left(2\right)}{3}+\frac{2}{\sqrt{3}}arctan\left(\frac{1}{\sqrt{3}}\right)=\frac{ln\left(2\right)}{3}+\frac{2}{\sqrt{3}}\frac{\pi }{6}=\frac{ln\left(2\right)}{3}+\frac{\pi }{3\sqrt{3}}\Rightarrow$$$${lim}_{n\to +\mathrm{\infty }}{S}_n=\frac{2ln\left(2\right)}{3}+\frac{\pi }{3\sqrt{3}}.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com