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Question Number 106366 by 175mohamed last updated on 04/Aug/20

Answered by mathmax by abdo last updated on 05/Aug/20

$$\mathrm{cosx}=1-\frac{{\mathrm{x}}^2}{2}+\frac{{\mathrm{x}}^4}{4!}+...\Rightarrow 1-\frac{{\mathrm{x}}^2}{2}⩽\mathrm{cosx}⩽1\Rightarrow$$$$1-\frac{1}{2}\left(\frac{1}{2^{\mathrm{k}}}\right)⩽\cos \left(\frac{1}{2^{\mathrm{k}}}\right)⩽1\Rightarrow \sum _{\mathrm{k}=1}^{2^{\mathrm{n}}}(1-\frac{1}{2^{\mathrm{k}+1}})⩽\sum _{\mathrm{k}=1}^{2^{\mathrm{n}}}\cos \left(\frac{1}{2^{\mathrm{k}}}\right)⩽2^{\mathrm{n}}\Rightarrow$$$$2^{\mathrm{n}}-\sum _{\mathrm{k}=1}^{2^{\mathrm{n}}}\frac{1}{2^{\mathrm{k}+1}}⩽\sum _{\mathrm{k}=1}^{2^{\mathrm{n}}}\cos \left(\frac{1}{2^{\mathrm{k}}}\right)⩽2^{\mathrm{n}}\mathrm{but}$$$$\sum _{\mathrm{k}=1}^{2^{\mathrm{n}}}\frac{1}{2^{\mathrm{k}+1}}=\sum _{\mathrm{k}=2}^{2^{\mathrm{n}}+1}\frac{1}{2^{\mathrm{k}}}=\sum _{\mathrm{k}=0}^{2^{\mathrm{n}}+1}{\left(\frac{1}{2}\right)}^{\mathrm{k}}-\frac{3}{2}=\frac{1-{\left(\frac{1}{2}\right)}^{2^{\mathrm{n}}+2}}{\frac{1}{2}}-\frac{3}{2}$$$$=2-\frac{2}{2^{2^{\mathrm{n}}+2}}-\frac{3}{2}=\frac{1}{2}-\frac{1}{2^{2^{\mathrm{n}}+1}}\Rightarrow$$$$2^{\mathrm{n}}-\frac{1}{2}-\frac{1}{2^{2^{\mathrm{n}}+1}}⩽\sum _{\mathrm{k}=1}^{2^{\mathrm{n}}}\cos \left(\frac{1}{2^{\mathrm{k}}}\right)⩽2^{\mathrm{n}}\Rightarrow$$$$1-\frac{1}{2^{\mathrm{n}+1}}-\frac{1}{2^{2^{\mathrm{n}}+1}.2^{\mathrm{n}}}⩽\frac{1}{2^{\mathrm{n}}}\sum _{\mathrm{k}=1}^{2^{\mathrm{n}}}\cos \left(\frac{1}{2^{\mathrm{k}}}\right)⩽1\Rightarrow$$$${\lim }_{\mathrm{n}\to +\mathrm{\infty }}\frac{1}{2^{\mathrm{n}}}\sum _{\mathrm{k}=1}^{2^{\mathrm{n}}}\cos \left(\frac{1}{2^{\mathrm{k}}}\right)=1$$

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