Question Number 113897 by Lordose last updated on 16/Sep/20

Answered by mathmax by abdo last updated on 17/Sep/20

let u_n =(3^n /(√((3n−2)2^n ))) ⇒u_n =((((3/(√2)))^n )/(√(3n−2)))  (n≥1) ⇒  (u_(n+1) /u_n ) =((((3/(√2)))^(n+1) )/(√(3n+1)))×((√(3n−2))/(((3/(√2)))^n )) =(√((3n−2)/(3n+1)))×((3/(√2)))  and lim_(n→+∞)  (u_(n+1) /u_n )  =(3/(√2))  so for ∣x∣ <(3/(√2))  the serie converges and for ∣x∣≥(3/(√2))  the serie diverges finally I_c =]−(3/(√2)),(3/(√2))[