Montrer-que-la-suite-de-finie-par-u-n-k-1-n-x-k-k-est-une-suite-de-Cauchy-

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

Montrer que la suite d \overset{´}{e f i n i e} par

u_{n} = \underset{k = 1}{\sum^{n}} \frac{x^{k}}{k} est une suite de Cauchy .

Answered by alcohol last updated on 21/Nov/21

l e t ε > 0 f i n d N_{ε} s . t \forall n, m \in N : n ⩾ m a n d n, m > N_{ε}

\Rightarrow ∣ U_{n} - U_{m} ∣ < ε

U_{n} = \underset{k = 1}{\sum^{n}} \frac{x^{k}}{k}; U_{m} = \underset{k = 1}{\sum^{m}} \frac{x^{k}}{k}

∣ U_{n} - U_{m} ∣=∣ \underset{k = 1}{\sum^{n}} \frac{x^{k}}{k} - \underset{k = 1}{\sum^{m}} \frac{x^{k}}{k} ∣

l e t n = m + 1

\Rightarrow ∣ U_{n} - U_{m} ∣=∣ \underset{k = 1}{\sum^{m + 1}} \frac{x^{k}}{k} - \underset{k = 1}{\sum^{m}} \frac{x^{k}}{k} ∣

\Rightarrow ∣ U_{n} - U_{m} ∣ = ∣ (\underset{k = 1}{\sum^{m}} \frac{x^{k}}{k} + \frac{x^{m + 1}}{m + 1}) - \underset{k = 1}{\sum^{m}} \frac{x^{k}}{k} ∣

\Rightarrow ∣ U_{n} - U_{m} ∣=∣ \frac{x^{m + 1}}{m + 1} ∣

∣ x ∣< 1 \Rightarrow ∣ x^{m + 1} ∣< 1 \Rightarrow ∣ \frac{x^{m + 1}}{m + 1} ∣< \frac{1}{m + 1} < m + 1

t a k e ε = m + 1 \Rightarrow m = ε - 1

\Rightarrow N_{ε} = E (ε - 1) + 2021

Commented by alcohol last updated on 21/Nov/21

w h a t u a s k e d

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

What are you doing, man ?

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

Thank you . I just had some doubts