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Question Number 132098 by rs4089 last updated on 11/Feb/21

Answered by Ar Brandon last updated on 11/Feb/21

$${\mathrm{S}}_{\mathrm{n}}=\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\frac{\mathrm{n}}{{\mathrm{n}}^2+\mathrm{kn}+{\mathrm{k}}^2}$$$$\text{Double subscripts: use braces to clarify}$$$$={\int }_0^1\frac{\mathrm{dx}}{{\mathrm{x}}^2+\mathrm{x}+1}={\int }_0^1\frac{\mathrm{dx}}{{(\mathrm{x}+\frac{1}{2})}^2+\frac{3}{4}}=\frac{2}{\sqrt{3}}{\left[{\tan }^{-1}\left(\frac{2\mathrm{x}+1}{\sqrt{3}}\right)\right]}_0^1$$$$=\frac{2}{\sqrt{3}}(\frac{\pi }{3}-\frac{\pi }{6})=\frac{\pi }{3\sqrt{3}}$$

Commented by Ar Brandon last updated on 11/Feb/21

$$\mathrm{For}\mathrm{n}=1$$$${\mathrm{S}}_1=\frac{1}{1+1+1}=\frac{1}{3}<\frac{\pi }{3\sqrt{3}}$$$${\mathrm{T}}_1=\frac{1}{1}>\frac{\pi }{3\sqrt{3}}$$

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