Question Number 179919 by mnjuly1970 last updated on 04/Nov/22
$$ \\ $$$$\:\:\:\:\:\:\:\mathrm{Evaluate}\: \\ $$$$\:\:\:\:\:\:\Omega\:=\:\mathrm{lim}_{\:{n}\rightarrow\infty} \left(\:{n}−\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{cos}\:\left(\:\frac{\:\sqrt{{k}}}{\:{n}}\:\:\right)\:\right)\:=?\:\:\:\:\:\: \\ $$$$\:\:\:\:\: \\ $$
Answered by mindispower last updated on 05/Nov/22
$${cos}\left({x}\right)=\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{sin}\left({c}\right).\frac{{x}^{\mathrm{3}} }{\mathrm{6}},\forall{x}\in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right]\exists{c}\in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right] \\ $$$$\:\:\:\:\:\:\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\leqslant{cos}\left({x}\right)\leqslant\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{6}} \\ $$$${S}=\Sigma{cos}\left(\frac{\sqrt{{k}}}{{n}}\right) \\ $$$$\Sigma\left(\mathrm{1}−\frac{{k}}{\mathrm{2}{n}^{\mathrm{2}} }\right)\leqslant{S}\leqslant\Sigma\left(\mathrm{1}−\frac{{k}}{\mathrm{2}{n}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{6}{n}^{\frac{\mathrm{3}}{\mathrm{2}}} }.\sqrt{\frac{{k}^{\mathrm{3}} }{{n}^{\mathrm{3}} }}\right) \\ $$$$\frac{{k}}{{n}}\leqslant\mathrm{1} \\ $$$${n}−\frac{\mathrm{1}}{\mathrm{4}{n}^{\mathrm{2}} }{n}\left({n}+\mathrm{1}\right)\leqslant{S}\leqslant{n}−\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{4}{n}^{\mathrm{2}} }+\frac{\mathrm{1}}{{n}^{\frac{\mathrm{3}}{\mathrm{2}}} \mathrm{6}}\sqrt{\frac{{k}^{\mathrm{3}} }{{n}^{\mathrm{3}} }}\leqslant \\ $$$${n}−\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{4}{n}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{6}\sqrt{\boldsymbol{{n}}}}\rightarrow \\ $$$$\:\:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{4}{n}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{6}\sqrt{{n}}}\rightarrow\frac{\mathrm{1}}{\mathrm{4}}\leqslant\left({n}−\Sigma{cos}\left(\frac{\sqrt{{k}}}{{n}}\right)\right)\leqslant\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{4}{n}^{\mathrm{2}} }\rightarrow\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{n}−\Sigma{cos}\frac{\sqrt{{k}}}{{n}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 05/Nov/22
$$\:\:\:{thanks}\:{alot}\:\:\: \\ $$$$\:\:{welcom}\:{again}\:{sir} \\ $$