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Question Number 97985 by abdomathmax last updated on 10/Jun/20

$$\mathrm{let}{\mathrm{S}}_{\mathrm{n}}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\sqrt{{\mathrm{n}}^2+2\mathrm{kn}}}$$$$\mathrm{find}{\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{S}}_{\mathrm{n}}$$

Answered by maths mind last updated on 11/Jun/20

$$S_n=\frac{1}{n}\sum _{1⩽k⩽n}\frac{1}{\sqrt{1+\frac{2k}{n}}}$$$$\underset{n\to \mathrm{\infty }}{\lim }{S}_n={\int }_0^1\frac{1}{\sqrt{1+2x}}-1$$$$={\left[2\sqrt{1+2x}\right]}_0^1-1$$$$=2\sqrt{3}-3$$

Answered by mathmax by abdo last updated on 12/Jun/20

$${\mathrm{S}}_{\mathrm{n}}=\frac{1}{\mathrm{n}}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\sqrt{1+2\frac{\mathrm{k}}{\mathrm{n}}}}\mathrm{so}{\mathrm{S}}_{\mathrm{n}}\mathrm{is}\mathrm{a}\mathrm{Rieman}\mathrm{sum}\mathrm{and}$$$${\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{S}}_{\mathrm{n}}={\int }_0^1\frac{\mathrm{dx}}{\sqrt{1+2\mathrm{x}}}={\left[\sqrt{1+2\mathrm{x}}\right]}_0^1=\sqrt{3}-1$$

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