Question Number 33699 by math khazana by abdo last updated on 22/Apr/18
![let S(x)=Σ_(n=0) ^∞ f_n (x) with f_n (x)= (((−1)^n )/(n!(x+n))) x∈]0,+∞[ 1) prove that S id defined .calculate S(1) and prove that ∀x>0 xS(x) −S(x+1) =(1/e) 2) prove that S is C^∞ on R^(+∗) 3) prove that S(x) ∼ (1/x) (x→0^+ ) .](https://www.tinkutara.com/question/Q33699.png)
Commented by math khazana by abdo last updated on 29/Apr/18
![we have ∣f_n (x)∣= (1/(n!(x+n))) ≤ (1/(n(n!))) but the seSrie Σ (1/(n(n!))) is convergent so S is defined on]0,+∞[ S(1)=Σ_(n=0) ^∞ f_n (1) =Σ_(n=0) ^∞ (((−1)^n )/(n!(n+1))) let w(x) = Σ_(n=0) ^∞ (((−1)^n )/(n!(n+1))) x^(n+1) w^′ (x) = Σ_(n=0) ^∞ (((−1)^n x^n )/(n!)) = Σ_(n=0) ^∞ (((−x)^n )/(n!)) =e^(−x) ⇒ w(x)=−e^(−x) +λ we have w(0)=0 =−1 +λ ⇒ λ=1 ⇒w(x)=1−e^(−x) S(1)=w(1)= 1−(1/e) . xS(x) −S(x+1)=Σ_(n=0) ^∞ ((x(−1)^n )/(n!(x+n))) −Σ_(n=0) ^∞ (((−1)^n )/(n!(x+n+1))) =1+Σ_(n=1) ^∞ (((−1)^n )/(n!)) ( ((x+n −n)/(x+n))) −Σ_(n=0) ^∞ (((−1)^n )/(n!(x+n+1))) =Σ_(n=0) ^∞ (((−1)^n )/(n!)) −Σ_(n=1) ^∞ (((−1)^n )/((n−1)!(x+n))) −Σ_(n=0) ^∞ (((−1)^n )/(n!(x+n+1))) = (1/e) −Σ_(n=0) ^∞ (((−1)^(n+1) )/(n!(x+n +1))) −Σ_(n=0) ^∞ (((−1)^n )/(n!(x+n+1))) =(1/e) ⇒ xS(x) −S(x+1) =(1/e)](https://www.tinkutara.com/question/Q34015.png)
Commented by math khazana by abdo last updated on 29/Apr/18
![2) due to uniform convergence and f_n are C^∞ S will be C^ on ]0,+∞[ and S^((p)) (x) =Σ_(n=0) ^∞ f_n ^((p)) (x)= Σ_(n=0) ^∞ (((−1)^n )/(n!)) (((−1)^p p!)/((x+n)^(p+1) )) .](https://www.tinkutara.com/question/Q34016.png)
Commented by math khazana by abdo last updated on 29/Apr/18
