let x_0 =1 and x_(n+1) =ln(e^x_n −x_n ) 1) prove that x_n →0 2)prove that Σ x


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Question Number 99839 by mathmax by abdo last updated on 23/Jun/20

$let x_{0} = 1 and x_{n + 1} = \ln (e^{x_{n}} - x_{n})$ $1) prove that x_{n} \to 0$ $2) prove that Σ x_{n} converges and ddyermine its sum$
Commented by bachamohamed last updated on 23/Jun/20

$d o y o u h a v e t h e s o l u t i o n ? s p e a d t h e s o l u t i o n f o r u s .$
Commented by abdomathmax last updated on 23/Jun/20

$you must take a try sir first . . .$
Commented by bachamohamed last updated on 23/Jun/20

$yes i tried and did not succed$ $we await your anser thank you sir$
Commented by mathmax by abdo last updated on 23/Jun/20

$nevermind we are here for help . . .$
	Answered by maths mind last updated on 23/Jun/20

	$1)$ $s h i w x_{n} c v$ $l e t s h o w x_{n} ⩾ 0$ $w e h a v e f (x) = e^{x} - x - 1 i s i n c r e a s i n g f o r x ⩾ 0$ $1 ⩾ x_{0} = 1 ⩾ 0$ $s u p o s e ⩾ 1 x_{n} ⩾ 0 \Rightarrow⩾ f (1) ⩾ f (x_{n}) ⩾ f (0) = 0 \Rightarrow e^{x_{n}} - x_{n} ⩾ 1 \Rightarrow l n (e^{x_{n}} - x_{n}) ⩾ 0$ $f (1) = e^{1} - 1 - 1 ⩽ 1 \Rightarrow x_{n + 1} ⩽ 1$ $\Rightarrow 1 ⩾ x_{n + 1} ⩾ 0 \Rightarrow b y i n d u c t i o n \forall n \in N 1 ⩾ x_{n} ⩾ 0$ $x_{n + 1} - x_{n} = l n (e^{x_{n}} - x_{n}) - x_{n} = l n (e^{x_{n}} - x_{n}) - l n (e^{x_{n}})$ $s i n c e e^{x_{n}} - x_{n} ⩽ e^{x n} \Rightarrow x_{n + 1} ⩽ x_{n}$ $\Rightarrow x_{n} i s b o u n d e d a n d d e a c r e a s e \Rightarrow c v$ $l = \lim_{n \to \infty} X_{n} \Rightarrow l n (e^{l} - l) = l \Rightarrow e^{l} - l = e^{l} \Rightarrow l = 0$ $x_{n + 1} = l n (e^{x_{n}} - x_{n})$ $n \to \infty x_{n} \to 0$ $e^{x_{n}} = 1 + x_{n} + \frac{{(x_{n})}^{2}}{2} + 0 (x_{n}^{2}) \Rightarrow e^{x n} - x_{n} = 1 + \frac{x_{n}^{2}}{2}$ $x_{n + 1} = l n (1 + \frac{x_{n}^{2}}{2} + o (x_{n}^{2}))$ $= \frac{x_{n}^{2}}{2} + 0 (x_{n}^{2}) \Rightarrow \exists c \in] 0, 1 [s u c h$ $\Rightarrow x_{n + 1} ⩽ {c x}_{n}^{2} = ({c x}_{n}) x_{n}$ $s i n c e x_{n} \to 0 \Rightarrow \exists N \forall n ⩾ N x_{n} < \frac{1}{2}$ $\Rightarrow x_{n + 1} ⩽ \frac{c}{2} x_{n} \Rightarrow x_{n} ⩽ {(\frac{c}{2})}^{n - N}, \forall n ⩾ N$ $\Rightarrow Σ x_{n} = \sum_{n ⩽ N} x_{n} + \sum_{n > N} x_{n}$ $f i r s t f i n i t e 2 n d b y c o m p a r a i s o n w i t h e c v g e o e m t r y s e r i e$ $\Rightarrow Σ x_{n} C v$ $i t S u m i w o r c k i n g o n i t$
	Commented by abdomathmax last updated on 23/Jun/20

	$thanks sir .$
	Answered by abdomathmax last updated on 23/Jun/20

	$2) let complete the solution we have$ $x_{n} \to 0 (n \to \infty) and e^{x_{n + 1}} = e^{x_{n}} - x_{n} \Rightarrow$ $x_{n} = e^{x_{n}} - e^{x_{n + 1}} let S_{n} = \sum_{k = 1}^{n} x_{k} \Rightarrow$ $S_{n} = \sum_{k = 0}^{n} (e^{x_{k}} - e^{x_{k + 1}}) = e^{x_{0}} - e^{x_{1}} + e^{x_{1}} - e^{x_{2}} + . . . .$ $+ e^{x_{n}} - e^{x_{n + 1}} = e - e^{x_{n + 1}} \to e - 1 \Rightarrow$ $\sum_{n = 0}^{\infty} x^{n} = e - 1$
	Commented by abdomathmax last updated on 23/Jun/20

	$\sum_{n = 0}^{\infty} x_{n} = e - 1$
	Commented by bachamohamed last updated on 23/Jun/20

	${thank}^{'} s sir$
	Commented by mathmax by abdo last updated on 23/Jun/20

	$you are welcome .$
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