- #1
eljose
- 492
- 0
let,s suppose that we have the limit with n tending to infinity:
[tex]\frac{f(n)}{g(n)}=1 [/tex] then i suppose that for n tending to infinity we should get:
[tex]f(n)\rightarrow{g(n)}[/tex] or what is the same the function f(n) diverges as g(n) as an special case:
[tex]\pi(n)\rightarrow{n/ln(n)} [/tex] where Pi is the prime number counting function in number theory
[tex]\frac{f(n)}{g(n)}=1 [/tex] then i suppose that for n tending to infinity we should get:
[tex]f(n)\rightarrow{g(n)}[/tex] or what is the same the function f(n) diverges as g(n) as an special case:
[tex]\pi(n)\rightarrow{n/ln(n)} [/tex] where Pi is the prime number counting function in number theory