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Question Number 129868 by Bird last updated on 20/Jan/21

$$find{lim}_{n\to +\mathrm{\infty }}\frac{1}{n}\sum _{k=0}^{n-1}\frac{k}{\sqrt{4n^2-k^2}}$$

Answered by Ar Brandon last updated on 20/Jan/21

$${\mathrm{S}}_{\mathrm{\infty }}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}-1}}\frac{\mathrm{k}}{\mathrm{n}\sqrt{4-\frac{{\mathrm{k}}^2}{{\mathrm{n}}^2}}}$$$$={\int }_0^1\frac{\mathrm{x}}{\sqrt{4-{\mathrm{x}}^2}}\mathrm{dx}={[-\sqrt{4-{\mathrm{x}}^2}]}_0^1$$$$=2-\sqrt{3}$$

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