I am struggling in evaluating the maximum of a function, for which I need an asymptotic, which I know how it is behaving. So I am trying to avoid the calculation of the maximum.
More precisely, I have the following setup: I have $f(n)$ and $g(n)$, with $f(n) \to \infty$, $g(n) \to 0$ and $f(n) g(n) \to 0$.
Now, I am wondering whether the following equality holds:
$\limsup_{n \to \infty} ( f(n) \max_{k \geq n} g(k) ) = \limsup_{n \to \infty} ( f(n) g(n) )$
My argument is the following: the maximum of $g$ is taken over the set $ \{ n, n+1, \ldots \} $. W.l.o.g let assume that the maximum of $g$ is taken at $\tilde{k}:=n+a$ with $a \in \mathbb{N}$, then
$\limsup_{n \to \infty} ( f(n) \max_{k \geq n} g(k) ) = \limsup_{n \to \infty} ( f(n) g(n+a) ) = \limsup_{n \to \infty} ( f(n) g(n) )$
Is that correct?
Any hint would be appreciated.
First, note that $f(n)g(n)\to 0 \implies \limsup_{n \to \infty} ( f(n) g(n) ) = 0.$
Here is a counter-example to the proposition in the question:
$ f(n)= \begin{cases} \log(n)&\text{if}\, n\text{ odd }\\ n&\text{if}\, n\text{ even}\\ \end{cases} $
$ g(n)= \begin{cases} 1/n&\text{if}\, n\text{ odd }\\ 2^{-n}&\text{if}\, n\text{ even}\\ \end{cases} $
Then we do have $f(n)\to\infty,\ g(n)\to 0,$ and $f(n)g(n)\to 0.$
But $\limsup_{n \to \infty} ( f(n) \max_{k \geq n} g(k) ) = 1\neq 0.$