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 34433 by abdo mathsup 649 cc last updated on 06/May/18

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

Commented by math khazana by abdo last updated on 06/May/18

$$letput{S}_n=\frac{1}{n}\sum _{k=1}^{n-1}\sqrt{\frac{n+k}{n-k}}$$$$S_n=\frac{1}{n}\sum _{k=1}^{n-1}\sqrt{\frac{1+\frac{k}{n}}{1-\frac{k}{n}}}\Rightarrow$$$${lim}_{n\to +\mathrm{\infty }}{S}_n={\int }_0^1\sqrt{\frac{1+x}{1-x}}dxchangement$$$$x=costgive{\int }_0^1\sqrt{\frac{1+x}{1-x}}dx=-{\int }_{\frac{\pi }{2}}^0\sqrt{\frac{{cos}^2\left(\frac{t}{2}\right)}{{sin}^2\left(\frac{t}{2}\right)}}sintdt$$$$={\int }_0^{\frac{\pi }{2}}cotan\left(\frac{t}{2}\right)sintdt$$$$={\int }_0^{\frac{\pi }{2}}\frac{cos\left(\frac{t}{2}\right)}{sin\left(\frac{t}{2}\right)}2sin\left(\frac{t}{2}\right)cos\left(\frac{t}{2}\right)dt$$$$={\int }_0^{\frac{\pi }{2}}2{cos}^2\left(\frac{t}{2}\right)dt={\int }_0^{\frac{\pi }{2}}(1+cost)dt$$$$=\frac{\pi }{2}+{\left[sint\right]}_0^{\frac{\pi }{2}}=1+\frac{\pi }{2}$$$$★{lim}_{n\to +\mathrm{\infty }}{S}_n=1+\frac{\pi }{2}★$$

Answered by arnabmaiti550@gmail.com last updated on 06/May/18

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}-1}}\sqrt{\frac{\mathrm{n}+\mathrm{k}}{\mathrm{n}-\mathrm{k}}}$$$$=\underset{\mathrm{h}\to 0}{\lim }\mathrm{h}\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}-1}}\sqrt{\frac{1/\mathrm{h}+\mathrm{k}}{1/\mathrm{h}-\mathrm{k}}}$$$$=\underset{\mathrm{h}\to 0}{\lim }\mathrm{h}\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}-1}}\sqrt{\frac{1+\mathrm{kh}}{1-\mathrm{kh}}}\mathrm{as}\mathrm{h}\to 0\mathrm{h}\ne 0$$$$=\underset{\mathrm{h}\to 0}{\lim }\mathrm{h}[\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}-1}}\sqrt{\frac{1+\mathrm{kh}}{1-\mathrm{kh}}}-\sqrt{\frac{1+0.\mathrm{h}}{1-0.\mathrm{h}}}]$$$$=\underset{\mathrm{h}\to 0}{\lim }\mathrm{h}\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}-1}}\sqrt{\frac{1+\mathrm{kh}}{1-\mathrm{kh}}}-\underset{\mathrm{h}\to 0}{\lim }\mathrm{h}$$$$={\int }_0^1\sqrt{\frac{1+\mathrm{x}}{1-\mathrm{x}}}\mathrm{dx}-0$$$$={\int }_0^1\frac{1+\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}$$$$={\int }_0^1\frac{\mathrm{dx}}{\sqrt{1-{\mathrm{x}}^2}}+{\int }_0^1\frac{\mathrm{x}\mathrm{dx}}{\sqrt{1-{\mathrm{x}}^2}}$$$$={\left[{\sin }^{-1}\left(\mathrm{x}\right)\right]}_0^1+{\int }_0^{\frac{\pi }{2}}\frac{\sin \theta \cos \theta \mathrm{d}\theta }{\cos \theta }[\mathrm{put}\mathrm{x}=\sin \theta ]$$$$=\frac{\pi }{2}+{\int }_0^{\frac{\pi }{2}}\sin \theta \mathrm{d}\theta$$$$=\frac{\pi }{2}+{[-\cos \theta ]}_0^{\frac{\pi }{2}}$$$$=\frac{\pi }{2}+1$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com