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Question Number 155806 by SANOGO last updated on 05/Oct/21

Answered by puissant last updated on 05/Oct/21

$$\mathcal{L}=\underset{n\to \mathrm{\infty }}{\lim }\underset{k=1}{\sum ^{n-1}}\frac{1}{\sqrt{n^2-k^2}}$$$$=\underset{n\to \mathrm{\infty }}{\lim }\underset{k=0}{\sum ^{n-1}}\frac{1}{\sqrt{n^2-k^2}}-\frac{1}{\sqrt{n^2}}$$$$=\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}\underset{k=0}{\sum ^{n-1}}\frac{1}{\sqrt{1-{\left(\frac{k}{n}\right)}^2}}-\frac{1}{n}$$$$or\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}=0ilrestedonc:$$$$\mathcal{L}=\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}\underset{k=0}{\sum ^{n-1}}\frac{1}{\sqrt{1-{\left(\frac{k}{n}\right)}^2}}quiestunesomme$$$$deRiemannetdonnealors:$$$$\mathcal{L}={\int }_0^1\frac{1}{\sqrt{1-x^2}}dxIlexisteplusieurs$$$$methodesmaisallonspas-a-paspour$$$$comprendre.$$$$posonsx=sint\to dx=costdt$$$$\mathcal{L}={\int }_0^{\frac{\pi }{2}}\frac{costdt}{\sqrt{1-{sin}^{2}t}}={\int }_0^{\frac{\pi }{2}}dt=\frac{\pi }{2}-0=\frac{\pi }{2}..$$$$$$$$\therefore \because \mathcal{L}=\underset{n\to \mathrm{\infty }}{\lim }\underset{k=1}{\sum ^{n-1}}\frac{1}{\sqrt{n^2-k^2}}=\frac{\pi }{2}..$$$$$$$$............\mathcal{L}epuissant............$$

Commented by SANOGO last updated on 05/Oct/21

$$mercibien$$

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