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Question Number 176589 by Ar Brandon last updated on 22/Sep/22

Answered by Peace last updated on 23/Sep/22

$$\mid u_n\left(z\right)\mid =\frac{\mid z{\mid }^n}{\mid 1+z^{2n+1}\mid }⩽\frac{\mid z{\mid }^n}{1-\mid z{\mid }^{2n+1}},\mid z\mid <1$$$$\mathrm{\exists }n\in \mathbb{N}\mid \mathrm{\forall }m⩾n\mid z\mid <\frac{1}{2}\Rightarrow .\mid U_m\mid ⩽2\mid z{\mid }^m..\mathrm{\Sigma }{U}_n.Cv.car\mathrm{\Sigma }\mid u_n\mid cv$$$$U_n\left(z\right)=\frac{\frac{1}{z^{n+1}}}{\frac{1}{z^{2n+1}}+1}=\frac{U_n\left(\frac{1}{z}\right)}{z}\Rightarrow \mid z\mid >1MemeRaisonnementMarche$$$$\left(2\right)\frac{z^n}{1+z^{2n+1}}=\sum _{k⩾0}{z}^n{(-z^{2n+1})}^k;\mathrm{\forall }\mid z\mid <1$$$$SoitS=\mathrm{\Sigma }{U}_n,S\left(z\right)=\sum _{n,k⩾0}{z}^n{(-z^{2n+1})}^k=\sum _{k⩾0}{(-z)}^k\sum _{n⩾0}{z}^{n(1+2k)}$$$$\sum _k\sum _n=\sum _n\sum _k.....\left(OnaCvAbsolut\right)$$$$\sum _{n⩾0}{z}^{n(1+2k)}=\frac{1}{1-z^{1+2k}}$$$$S=\sum _{k⩾0}\frac{{(-z)}^k}{1-z^{1+2k}}=U(-z)$$$$\Rightarrow s\left(z\right)=s(-z),\mathrm{\forall }\mid z\mid <1Spair/pour\mid z\mid >1$$$$S\left(z\right)=\frac{S\left(\frac{1}{z}\right)}{z},s(-z)=\frac{s(-\frac{1}{z})}{-z}=-\frac{s\left(\frac{1}{z}\right)}{z}=-s\left(z\right)Impaire$$$$$$

Commented by Tawa11 last updated on 23/Sep/22

$$\mathrm{Great}\mathrm{sirs}$$

Commented by Ar Brandon last updated on 23/Sep/22

Thanks, comrade !

Commented by Peace last updated on 25/Sep/22

$$withepleasurthank$$$$avecplaisirbonnejournee$$

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