= Solution
Put $a=1+\epsilon$ and, for $m\geq2$, define
$$
I_m(\epsilon)=\int_0^\infty(1+\epsilon t)^{-m}e^{-at}\,dt.
$$
One <integration by parts> gives
$$
I_m(\epsilon)=\frac1a-\frac{m\epsilon}{a}I_{m+1}(\epsilon).
$$
Starting with $I=I_2$ and iterating $N+1$ times yields
$$
I(\epsilon)
=\sum_{n=0}^N
(-1)^n(n+1)!\frac{\epsilon^n}{(1+\epsilon)^{n+1}}
+R_N(\epsilon),
$$
where
$$
R_N(\epsilon)
=(-1)^{N+1}(N+2)!
\frac{\epsilon^{N+1}}{(1+\epsilon)^{N+1}}I_{N+3}(\epsilon).
$$
Since $0<I_{N+3}(\epsilon)\leq(1+\epsilon)^{-1}$, the remainder is $O(\epsilon^{N+1})$, which is little-$o$ of the last retained term. Therefore
$$
\boxed{
I(\epsilon)\sim
\sum_{n=0}^{\infty}
(-1)^n(n+1)!\frac{\epsilon^n}{(1+\epsilon)^{n+1}}
}
\qquad(\epsilon\to0^+).
$$
The displayed terms form an <asymptotic sequence> because the ratio of successive magnitudes tends to zero.
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