Răspuns:
∑₀⁰⁰3(a/9)ⁿ=3∑(a/9)ⁿ
a)Seria e covergenta daca (a/9)ⁿ->0 adica a/9∈(-1,1) adica a∈(-9,9)
Pt a=5 seria este
3∑(5/9)ⁿ
b)Calculezi suma sirurilor partiale
Sn =(5/9)⁰+(5/9)¹+(5/9)²+...+(5/9)ⁿ
=1+(5/9)¹+(5/9)²+...+(5/9)ⁿ
Ai o proogresie geo,metrica de ratie 5/9
Sn=(5/9)*[1-(5/9)ⁿ⁺₁]/(1-5/9)=
(5/9)[1-5/9)ⁿ⁺¹]/(4/9)=(5/4)[[1-(5/9)ⁿ⁺¹
Lim Sₙ=5/4=>
3∑(5/9)ⁿ=3*5/4=15/4
Explicație pas cu pas:
