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Question Number 179919 by mnjuly1970 last updated on 04/Nov/22

$$$$$$\mathrm{Evaluate}$$$$\mathrm{\Omega }={\lim }_{n\to \mathrm{\infty }}(n-\underset{k=1}{\sum ^n}\cos \left(\frac{\sqrt{k}}{n}\right))=?$$$$$$

Answered by mindispower last updated on 05/Nov/22

$$cos\left(x\right)=1-\frac{x^2}{2}+sin\left(c\right).\frac{x^3}{6},\mathrm{\forall }x\in [0,\frac{\pi }{2}]\mathrm{\exists }c\in [0,\frac{\pi }{2}]$$$$1-\frac{x^2}{2}⩽cos\left(x\right)⩽1-\frac{x^2}{2}+\frac{x^3}{6}$$$$S=\mathrm{\Sigma }cos\left(\frac{\sqrt{k}}{n}\right)$$$$\mathrm{\Sigma }(1-\frac{k}{2n^2})⩽S⩽\mathrm{\Sigma }(1-\frac{k}{2n^2}+\frac{1}{6n^{\frac{3}{2}}}.\sqrt{\frac{k^3}{n^3}})$$$$\frac{k}{n}⩽1$$$$n-\frac{1}{4n^2}n(n+1)⩽S⩽n-\frac{n(n+1)}{4n^2}+\frac{1}{n^{\frac{3}{2}}6}\sqrt{\frac{k^3}{n^3}}⩽$$$$n-\frac{n(n+1)}{4n^2}+\frac{1}{6\sqrt{\mathit{n}}}\to$$$$\frac{n(n+1)}{4n^2}-\frac{1}{6\sqrt{n}}\to \frac{1}{4}⩽(n-\mathrm{\Sigma }cos\left(\frac{\sqrt{k}}{n}\right))⩽\frac{n(n+1)}{4n^2}\to \frac{1}{4}$$$$\underset{n\to \mathrm{\infty }}{\lim }n-\mathrm{\Sigma }cos\frac{\sqrt{k}}{n}=\frac{1}{4}$$$$$$$$$$

Commented by mnjuly1970 last updated on 05/Nov/22

$$thanksalot$$$$welcomagainsir$$

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