(3^n+4^n)/12^n=3^n/12^n+4^n/12^n=1/4^n+1/3^n∈(0,1)
Ssirul partial e marginit, deci seria e convergenta
Rescrii seria ca suma de 2 serii
∑1/4^n+∑1/3^n n={0 ,∞}
∑1/4^n=1+1/4+1/4^2+...+1/4^n= progresie geometrica cu ratia 1/4=(1/4^(n+1)-1)/(1/4-1)=-4*[(1/4(n+1)-1]/3
∑1/3^n=1+1/3+1/3^2+...+3^n=(3^n+1-1)/(1/3-1)=-3(3^(n+1)-1)/2