= Solution
Since
$$
\frac{x}{x-1}=\frac1{1-x^{-1}}
\sim\sum_{j=0}^\infty x^{-j},
$$
we have
$$
\boxed{
f(x)\sim x^{-1}+\sum_{n=1}^\infty x^{-n+1}e^{-x}
=\sum_{n=0}^\infty\phi_n(x)}.
$$
Thus $a_n=1$ for every $n\geq0$.
Explicitly,
$$
\frac f{\phi_0}
=1+\frac{x^2e^{-x}}{x-1}\longrightarrow1.
$$
For $n\geq1$, the remainder after terms $0,\ldots,n-1$ is
$$
e^{-x}\left(\frac1{1-x^{-1}}
-\sum_{j=0}^{n-2}x^{-j}\right)
=\frac{x^{-n+1}e^{-x}}{1-x^{-1}}.
$$
Its ratio to $\phi_n$ tends to one, verifying every coefficient formula.
Because $e^{-x}=o(x^{-N})$ for every $N$, the exponentially small part is invisible to the power <sequence> $\psi_n=x^{-n}$. Hence
$$
\boxed{f(x)\sim x^{-1}}
$$
with coefficient one at $\psi_1$ and all other power coefficients zero.
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