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

Commented by bachamohamed last updated on 23/Jun/20

Commented by abdomathmax last updated on 23/Jun/20

Commented by bachamohamed last updated on 23/Jun/20

Commented by mathmax by abdo last updated on 23/Jun/20

Answered by maths mind last updated on 23/Jun/20
![1) shiw x_n cv let show x_n ≥0 we havef(x)=e^x −x−1 is increasing for x≥0 1≥x_0 =1≥0 supose≥1 x_n ≥0⇒≥f(1)≥f(x_n )≥f(0)=0⇒e^x_n −x_n ≥1⇒ln(e^x_n −x_n )≥0 f(1)=e^1 −1−1≤1⇒x_(n+1) ≤1 ⇒1≥x_(n+1) ≥0⇒by induction ∀n∈N 1≥ x_n ≥0 x_(n+1) −x_n =ln(e^x_n −x_n )−x_n =ln(e^x_n −x_n )−ln(e^x_n ) since e^x_n −x_n ≤e^(xn) ⇒x_(n+1) ≤x_n ⇒x_n is bounded and deacrease ⇒cv l=lim_(n→∞) X_n ⇒ln(e^l −l)=l⇒e^l −l=e^l ⇒l=0 x_(n+1) =ln(e^x_n −x_n ) n→∞ x_n →0 e^x_n =1+x_n +(((x_n )^2 )/2)+0(x_n ^2 )⇒e^(xn) −x_n =1+(x_n ^2 /2) x_(n+1) =ln(1+(x_n ^2 /2)+o(x_n ^2 )) =(x_n ^2 /2)+0(x_n ^2 )⇒∃c∈]0,1[ such ⇒x_(n+1) ≤cx_n ^2 =(cx_n )x_n since x_n →0⇒∃N ∀n≥N x_n <(1/2) ⇒x_(n+1) ≤(c/2)x_n ⇒x_n ≤((c/2))^(n−N) ,∀n≥N ⇒Σx_n =Σ_(n≤N) x_n +Σ_(n>N) x_n first finite 2nd by comparaison withe cv geoemtry serie ⇒Σx_n Cv it Sum i worcking on it](https://www.tinkutara.com/question/Q99854.png)
Commented by abdomathmax last updated on 23/Jun/20

Answered by abdomathmax last updated on 23/Jun/20

Commented by abdomathmax last updated on 23/Jun/20

Commented by bachamohamed last updated on 23/Jun/20

Commented by mathmax by abdo last updated on 23/Jun/20
