Limit with n tending to infinity

[eljose]

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

 

[matt grime]

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

What is your question? What you;'ve written doesn't make sense.

Are you attempting to ask: if f and g are asymptotic then does f-g tend to zero? Of course not: we can make two functions that are asymptotic diverge absolutely as fast as you want.

 

[benorin]

As I recall, that \frac{f(n)}{g(n)}=1\mbox{ as }n\rightarrow\infty speaks of the asymptotic relationship between f and g, namely that f(n) \sim g(n) (ref. Asymptotic Notation).

 

[eljose]

My main question is related with proving (if true) the equality

\frac{\int_0^{n}dx(x^p)}{1^p+2^p+3^p+...+n^p}\rightarrow{1}

as n tends to infinity n\rightarrow{\infty}

 

[matt grime]

It certainly tends to a constant (for p an integer), as anyone can tell you, since the sum of the first p powers is a poly of degre p+1. Surely you can actually solve it, it's straightforward (especially from someone who has solved the RH amongst other things that snobbish mathematicians wont' acknowledge (you cry wolf and what do you expect?)), at least try simplifying the integral (ie doing it).