Hi guys I am trying to prove to myself that $\frac{1+e^{-t}}{1+t}$ and $\frac{1}{1+t}$ have the same asymptotic expansion as $t\rightarrow \infty$
Please correct me if I am wrong, but my thinking is to derive an expantion for each one of them let us call the expantions $e_1(t)$ and $e_2(t)$ then observe that $\lim_{t \rightarrow \infty} \frac{e_2(t)}{e_1(t)}= 1$. My question is this enough? And do my calculations look okay.
$\frac{1}{1+t}= \frac{1}{t}- \frac{1}{t^2}+ \frac{1}{t^3} -O(\frac{1}{t^4})$
Where we can call this $e_1(t)$. We do similarity for the other one
$\frac{1+e^{-t}}{1+t}=(1+e^{-t})\frac{1}{1+t} =(e^{-t}+1)\left[\frac{1}{t}- \frac{1}{t^2}+ \frac{1}{t^3} -O(\frac{1}{t^4})\right]$
which we can call $e_2$ then if we look at the
$\lim_{t \rightarrow \infty} \frac{e_2(t)}{e_1(t)}= 1$.
What do you guys think is this an exceptionable explanation?
No. To have "the same asymptotic expansion" is... to have the same asymptotic expansion. That is, the two asympotic expansions (mind you, in some prescribed sequence of functions - I guess it's $\{t^{-1},t^{-2}\cdots\}$ here) must be identical. That the ratio tends to one only shows that the functions have the same order of growth, which is a much weaker property.
BTW, in you resolution, $e_2$ is not an asymptotic expansion.
Actually, you already know the candidate a.e. of the second function... You need to prove that
$$ e(t)= \frac{1}{t}-\frac{1}{t^2}+\frac{1}{t^3}-\cdots$$
is indeed an a.e. of $$f_2(t)=\frac{1+e^{-t}}{1+t}$$
For this, you must show that
$$\frac{1}{t}-\frac{1}{t^2}+\cdots + (-1)^{n+1} \frac{1}{t^n} - f_2(t) = o( \frac{1}{t^n})$$