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

Montrer que la suite de^� finie par u_n =Σ_(k=1) ^n (x^k /k) est une suite de Cauchy.

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

Answered by alcohol last updated on 21/Nov/21

let ε>0 find N_ε s.t ∀n,m ∈N: n≥m and n,m>N_ε ⇒ ∣U_n −U_m ∣ <ε U_n =Σ_(k=1) ^n (x^k /k) ; U_m =Σ_(k=1) ^m (x^k /k) ∣U_n −U_m ∣=∣Σ_(k=1 ) ^n (x^k /k)−Σ_(k=1) ^m (x^k /k)∣ let n = m+1 ⇒ ∣U_n −U_m ∣=∣Σ_(k=1) ^(m+1) (x^k /k)−Σ_(k=1) ^m (x^k /k)∣ ⇒ ∣U_n −U_m ∣ = ∣(Σ_(k=1) ^m (x^k /k) + (x^(m+1) /(m+1)))−Σ_(k=1) ^m (x^k /k)∣ ⇒ ∣U_n −U_m ∣=∣(x^(m+1) /(m+1))∣ ∣x∣<1 ⇒ ∣x^(m+1) ∣<1 ⇒ ∣(x^(m+1) /(m+1))∣<(1/(m+1))<m+1 take ε = m+1 ⇒ m=ε−1 ⇒ N_ε =E(ε−1) + 2021

$${let}\:\varepsilon>\mathrm{0}\:{find}\:{N}_{\varepsilon} \:{s}.{t}\:\forall{n},{m}\:\in\mathbb{N}:\:{n}\geqslant{m}\:{and}\:{n},{m}>{N}_{\varepsilon} \\ $$$$\Rightarrow\:\mid{U}_{{n}} −{U}_{{m}} \mid\:<\varepsilon \\ $$$${U}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{x}^{{k}} }{{k}}\:;\:{U}_{{m}} =\underset{{k}=\mathrm{1}} {\overset{{m}} {\sum}}\frac{{x}^{{k}} }{{k}} \\ $$$$\mid{U}_{{n}} −{U}_{{m}} \mid=\mid\underset{{k}=\mathrm{1}\:} {\overset{{n}} {\sum}}\frac{{x}^{{k}} }{{k}}−\underset{{k}=\mathrm{1}} {\overset{{m}} {\sum}}\frac{{x}^{{k}} }{{k}}\mid \\ $$$${let}\:{n}\:=\:{m}+\mathrm{1} \\ $$$$\Rightarrow\:\mid{U}_{{n}} −{U}_{{m}} \mid=\mid\underset{{k}=\mathrm{1}} {\overset{{m}+\mathrm{1}} {\sum}}\frac{{x}^{{k}} }{{k}}−\underset{{k}=\mathrm{1}} {\overset{{m}} {\sum}}\frac{{x}^{{k}} }{{k}}\mid \\ $$$$\Rightarrow\:\mid{U}_{{n}} −{U}_{{m}} \mid\:=\:\mid\left(\underset{{k}=\mathrm{1}} {\overset{{m}} {\sum}}\frac{{x}^{{k}} }{{k}}\:+\:\frac{{x}^{{m}+\mathrm{1}} }{{m}+\mathrm{1}}\right)−\underset{{k}=\mathrm{1}} {\overset{{m}} {\sum}}\frac{{x}^{{k}} }{{k}}\mid \\ $$$$\Rightarrow\:\mid{U}_{{n}} −{U}_{{m}} \mid=\mid\frac{{x}^{{m}+\mathrm{1}} }{{m}+\mathrm{1}}\mid \\ $$$$\mid{x}\mid<\mathrm{1}\:\Rightarrow\:\mid{x}^{{m}+\mathrm{1}} \mid<\mathrm{1}\:\Rightarrow\:\mid\frac{{x}^{{m}+\mathrm{1}} }{{m}+\mathrm{1}}\mid<\frac{\mathrm{1}}{{m}+\mathrm{1}}<{m}+\mathrm{1} \\ $$$${take}\:\varepsilon\:=\:{m}+\mathrm{1}\:\Rightarrow\:{m}=\varepsilon−\mathrm{1} \\ $$$$\Rightarrow\:{N}_{\varepsilon} ={E}\left(\varepsilon−\mathrm{1}\right)\:+\:\mathrm{2021} \\ $$$$ \\ $$

Commented by alcohol last updated on 21/Nov/21

what u asked

$${what}\:{u}\:{asked} \\ $$

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

What are you doing, man ?

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

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

Thank you. I just had some doubts

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

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