Folosesti criteriu raportului Daca (an+1/an)<1 limita e 0 si daca (an+1/an)>1 an→∞
Fie an=(n^2016*2^n)/3^n
an+1=[(n+1)^2016*2^(n+1)/3^(n+1)
an+1/an=[(n+1)^2016*2^(n+1)/3^(n+1)]*[3^(n)/n^2016*2^n]=[(n+1)/n]^2016*[2^(n+1/2^n]*1/3>1 Pt ca fiecare [.] e supranunitara .Deci (an+1/an)>1=.>
an→∞ 1/3^n→0 atunci limita ta tinde la infinit
