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Question Number 120541 by 77731 last updated on 01/Nov/20

Answered by snipers237 last updated on 01/Nov/20

$$Usingcos\left(z\right)=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n{z}^{2n}}{\left(2n\right)!},1-cosz=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n{z}^{2n}}{\left(2n\right)!}$$$$\mid 1-cosz\mid ⩽\underset{n=1}{\sum ^{\mathrm{\infty }}}\mid \frac{{(-1)}^n{z}^{2n}}{\left(2n\right)!}\mid =\sum _{n=1}\frac{\mid z{\mid }^{2n}}{\left(2n\right)!}$$$$\mid 1-cosz\mid ⩽\mid z{\mid }^2\sum _{p=0}\frac{\mid z{\mid }^{2p}}{(2p+2)!}⩽\frac{\mid z{\mid }^2}{2}\underset{p=0}{\sum ^{\mathrm{\infty }}}\frac{\mid z{\mid }^{2p}}{\left(2p\right)!}cause(2p+2)⩾2$$$$Observesthat{e}^{\mid z\mid }=\underset{p=0}{\sum ^{\mathrm{\infty }}}\frac{\mid z{\mid }^{2p}}{\left(2p\right)!}+\underset{p=0}{\sum ^{\mathrm{\infty }}}\frac{\mid z{\mid }^{2p+1}}{(2p+1)!}⩾\sum _{p⩾0}\frac{\mid z{\mid }^{2p}}{\left(2p\right)!}$$$$$$$$LFYTC$$

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