The question was inspired by this discussion:
How to expand a function into a power series with negative powers?
I am interested in asymptotic behavior of a function at infinity:
f[r_]:=(0.04962 Exp[-2 r] (-1.000+r))/((0.06119+(Exp[-2 r])^(2/3))^2 r)
or (TeX)
$$ f(r)=\frac{0.04962 e^{-2 r} (r-1.000)}{\left(\left(e^{-2 r}\right)^{2/3}+0.06119\right)^2 r} $$
Tried Series[f[r],{r,0,10}] for expansion in negative powers at infinity (as suggested) and got:
$$ -\frac{0.04406}{r}+0.02146+0.02106 r+0.004405 r^2-0.001355 r^3-0.001205 r^4-0.0003607 r^5-\left(8.402\times 10^{-6}\right) r^6+0.00004149 r^7+0.00001982 r^8+\left(3.921\times 10^{-6}\right) r^9-\left(6.018\times 10^{-7}\right) r^{10}+O\left(r^{11}\right) $$
Seems like the function decays faster than $1/r^n$ and the expansion is meaningless. But what does the term $-\frac{0.04406}{r}$ mean then? The function is strictly positive at infinity and I am kinda confused by that. Does this mean that the asymptotic form of the function is something plus the term $-\frac{0.04406}{r}$ which effectively gives the observed behavior? Can someone clarify it? How can one explain the term $-\frac{0.04406}{r}$?

Seriescall. 3) In the answer you linked to, the series is developed at infinity, you use 0. – Sjoerd C. de Vries Jan 27 at 18:52Series[f[r],{r,0,10}]is equivalent toSeries[f[r],{1/r,Infinity,10}], isn't it? we can not use negative powers inSeries, though – molkee Jan 27 at 19:20