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Question Number 159796 by Ar Brandon last updated on 21/Nov/21

$$\mathrm{Montrer}\mathrm{que}\mathrm{la}\mathrm{suite}\mathrm{d}\acute{\mathrm{e}finie}\mathrm{par}$$$$u_n=\underset{k=1}{\sum ^n}\frac{x^k}{k}\mathrm{est}\mathrm{une}\mathrm{suite}\mathrm{de}\mathrm{Cauchy}.$$

Answered by alcohol last updated on 21/Nov/21

$$let\epsilon >0find{N}_{\epsilon }s.t\mathrm{\forall }n,m\in \mathbb{N}:n⩾mandn,m>N_{\epsilon }$$$$\Rightarrow \mid U_n-U_m\mid <\epsilon$$$$U_n=\underset{k=1}{\sum ^n}\frac{x^k}{k};U_m=\underset{k=1}{\sum ^m}\frac{x^k}{k}$$$$\mid U_n-U_m\mid =\mid \underset{k=1}{\sum ^n}\frac{x^k}{k}-\underset{k=1}{\sum ^m}\frac{x^k}{k}\mid$$$$letn=m+1$$$$\Rightarrow \mid U_n-U_m\mid =\mid \underset{k=1}{\sum ^{m+1}}\frac{x^k}{k}-\underset{k=1}{\sum ^m}\frac{x^k}{k}\mid$$$$\Rightarrow \mid U_n-U_m\mid =\mid (\underset{k=1}{\sum ^m}\frac{x^k}{k}+\frac{x^{m+1}}{m+1})-\underset{k=1}{\sum ^m}\frac{x^k}{k}\mid$$$$\Rightarrow \mid U_n-U_m\mid =\mid \frac{x^{m+1}}{m+1}\mid$$$$\mid x\mid <1\Rightarrow \mid x^{m+1}\mid <1\Rightarrow \mid \frac{x^{m+1}}{m+1}\mid <\frac{1}{m+1}<m+1$$$$take\epsilon =m+1\Rightarrow m=\epsilon -1$$$$\Rightarrow N_{\epsilon }=E(\epsilon -1)+2021$$$$$$

Commented by alcohol last updated on 21/Nov/21

$$whatuasked$$

Commented by Ar Brandon last updated on 21/Nov/21

$$\mathrm{What}\mathrm{are}\mathrm{you}\mathrm{doing},\mathrm{man}?$$

Commented by Ar Brandon last updated on 21/Nov/21

$$\mathrm{Thank}\mathrm{you}.\mathrm{I}\mathrm{just}\mathrm{had}\mathrm{some}\mathrm{doubts}$$

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