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(x)=1−(x^2 /2)+sin(c).(x^3 /6),∀x∈[0,(π/2)]∃c∈[0,(π/2)] 1−(x^2 /2)≤cos(x)≤1−(x^2 /2)+(x^3 /6) S=Σcos(((√k)/n)) Σ(1−(k/(2n^2 )))≤S≤Σ(1−(k/(2n^2 ))+(1/(6n^(3/2) )).(√(k^3 /n^3 ))) (k/n)≤1 n−(1/(4n^2 ))n(n+1)≤S≤n−((n(n+1))/(4n^2 ))+(1/(n^(3/2) 6))(√(k^3 /n^3 ))≤ n−((n(n+1))/(4n^2 ))+(1/(6(√n)))→ ((n(n+1))/(4n^2 ))−(1/(6(√n)))→(1/4)≤(n−Σcos(((√k)/n)))≤((n(n+1))/(4n^2 ))→(1/4) lim_(n→∞) n−Σcos((√k)/n)=(1/4)](https://www.tinkutara.com/question/Q179982.png)
$${cos}\left({x}\right)=\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{sin}\left({c}\right).\frac{{x}^{\mathrm{3}} }{\mathrm{6}},\forall{x}\in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right]\exists{c}\in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right] \\ $$$$\:\:\:\:\:\:\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\leqslant{cos}\left({x}\right)\leqslant\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{6}} \\ $$$${S}=\Sigma{cos}\left(\frac{\sqrt{{k}}}{{n}}\right) \\ $$$$\Sigma\left(\mathrm{1}−\frac{{k}}{\mathrm{2}{n}^{\mathrm{2}} }\right)\leqslant{S}\leqslant\Sigma\left(\mathrm{1}−\frac{{k}}{\mathrm{2}{n}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{6}{n}^{\frac{\mathrm{3}}{\mathrm{2}}} }.\sqrt{\frac{{k}^{\mathrm{3}} }{{n}^{\mathrm{3}} }}\right) \\ $$$$\frac{{k}}{{n}}\leqslant\mathrm{1} \\ $$$${n}−\frac{\mathrm{1}}{\mathrm{4}{n}^{\mathrm{2}} }{n}\left({n}+\mathrm{1}\right)\leqslant{S}\leqslant{n}−\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{4}{n}^{\mathrm{2}} }+\frac{\mathrm{1}}{{n}^{\frac{\mathrm{3}}{\mathrm{2}}} \mathrm{6}}\sqrt{\frac{{k}^{\mathrm{3}} }{{n}^{\mathrm{3}} }}\leqslant \\ $$$${n}−\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{4}{n}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{6}\sqrt{\boldsymbol{{n}}}}\rightarrow \\ $$$$\:\:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{4}{n}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{6}\sqrt{{n}}}\rightarrow\frac{\mathrm{1}}{\mathrm{4}}\leqslant\left({n}−\Sigma{cos}\left(\frac{\sqrt{{k}}}{{n}}\right)\right)\leqslant\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{4}{n}^{\mathrm{2}} }\rightarrow\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{n}−\Sigma{cos}\frac{\sqrt{{k}}}{{n}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 05/Nov/22
$$\:\:\:{thanks}\:{alot}\:\:\: \\ $$$$\:\:{welcom}\:{again}\:{sir} \\ $$