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Question Number 153384 by SANOGO last updated on 06/Sep/21

Answered by mindispower last updated on 06/Sep/21

$$\underset{n\to \mathrm{\infty }}{\lim }\mid \frac{U_{n+1}}{U_n}\mid =0$$$$\Rightarrow ϵ=\frac{1}{2},\mathrm{\exists }N\in \mathbb{N}\mathrm{\forall }n⩾N\mid \frac{U_{n+1}}{U_n}\mid ⩽\frac{1}{2}$$$$\Rightarrow \underset{k=N}{\prod ^{N+m}}\mid \frac{U_{k+1}}{U_k}\mid ⩽\underset{k=N}{\prod ^{m+N}}\left(\frac{1}{2}\right)$$$$\Rightarrow \mid U_{N+m+1}\mid ⩽U_N{\left(\frac{1}{2}\right)}^{m+1}$$$$\underset{m\to \mathrm{\infty }}{\lim }\mid U_{N+m+1}\mid ⩽U_N\underset{m\to \mathrm{\infty }}{\lim }{\left(\frac{1}{2}\right)}^{m+1}=0$$$$$$

Commented by SANOGO last updated on 07/Sep/21

$$mercibeaucoupchef$$

Answered by Jonathanwaweh last updated on 07/Sep/21

$$par{l}^{\prime }absurdesupposons\underset{n\to \mathrm{\infty }}{lim}\mid \frac{u_{n+1}}{u_n}\mid =0etlim\left(u_n\right)=l;(l\ne 0)$$$$alorspuisque\left(u_{n+1}\right)estunesous-suite\left(u_{n)}\right)ona{limu}_{n+1}=l$$$$ilvientlim\left(\frac{u_{n+1}}{u_n}\right)=\frac{{limu}_{n+1}}{{limu}_n}$$$$=\frac{l}{l}$$$$=1(absurdecarlim\mid \frac{U_{n+1}}{U_n}\mid =0\Rightarrow lim\left(\frac{u_{n+1}}{u_n}\right)=0)$$$$d^{\prime }ou....$$

Commented by SANOGO last updated on 07/Sep/21

$$mercibien$$

Commented by mindispower last updated on 23/Sep/21

$$jevousenPrie$$

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