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Question Number 111755 by mathmax by abdo last updated on 04/Sep/20

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

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

$$\frac{1}{n}\underset{n\to \mathrm{\infty }}{\lim }\underset{k=1}{\sum ^n}\sqrt{\frac{n-k}{n+k}}=\frac{1}{n}\underset{n\to \mathrm{\infty }}{\lim }\sqrt{\frac{1-\frac{k}{n}}{1+\frac{k}{n}}}={\int }_0^1\sqrt{\frac{1-x}{1+x}}dx$$$$={\int }_0^1\frac{1-x}{\sqrt{1-x^2}}$$$$={\int }_0^1\frac{1}{\sqrt{1-x^2}}+\frac{1}{2}{\int }_0^1\frac{-2x}{\sqrt{1-x^2}}$$$$={\left[{sin}^{-1}x\right]}_0^1+{\left[\sqrt{1-x^2}\right]}_0^1$$$$=\frac{\pi }{2}-1$$

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

$$\mathrm{let}{\mathrm{S}}_{\mathrm{n}}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\sqrt{\frac{\mathrm{n}-\mathrm{k}}{{\mathrm{n}}^3+{\mathrm{n}}^2\mathrm{k}}}\Rightarrow {\mathrm{S}}_{\mathrm{n}}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\sqrt{\frac{\mathrm{n}(1-\frac{\mathrm{k}}{\mathrm{n}})}{{\mathrm{n}}^3(1+\frac{\mathrm{k}}{\mathrm{n}})}}$$$$=\frac{1}{\mathrm{n}}\sum _{\mathrm{k}=1}^{\mathrm{n}}\sqrt{\frac{1-\frac{\mathrm{k}}{\mathrm{n}}}{1+\frac{\mathrm{k}}{\mathrm{n}}}}\to {\int }_0^1\sqrt{\frac{1-\mathrm{x}}{1+\mathrm{x}}}\mathrm{dx}\left({\mathrm{S}}_{\mathrm{n}}\mathrm{is}\mathrm{Rieman}\mathrm{sum}\right)$$$$\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{x}=\cos \theta \Rightarrow {\int }_0^1\sqrt{\frac{1-\mathrm{x}}{1+\mathrm{x}}}\mathrm{dx}$$$$={\int }_{\frac{\pi }{2}}^0\sqrt{\frac{{2\sin }^2\left(\frac{\theta }{2}\right)}{{2\cos }^2\left(\frac{\theta }{2}\right)}}(-\sin \theta )\mathrm{d}\theta =2{\int }_0^{\frac{\pi }{2}}\frac{\sin \left(\frac{\theta }{2}\right)}{\cos \left(\frac{\theta }{2}\right)}\sin \left(\frac{\theta }{2}\right)\cos \left(\frac{\theta }{2}\right)\mathrm{d}\theta$$$$={\int }_0^{\frac{\pi }{2}}{2\sin }^2\left(\frac{\theta }{2}\right)\mathrm{d}\theta ={\int }_0^{\frac{\pi }{2}}(1-\cos \theta )\mathrm{d}\theta =\frac{\pi }{2}-{\left[\sin \theta \right]}_0^{\frac{\pi }{2}}$$$$=\frac{\pi }{2}-1\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{S}}_{\mathrm{n}}=\frac{\pi }{2}-1$$

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