= Solution
An <asymptotic power series> about $a$ means that for every fixed $N\ge0$,
$$
f(x)-\sum_{k=0}^Nc_k(x-a)^k=o(|x-a|^N)
$$
as $x\to a$ through the specified domain. It makes no assertion that the infinite series converges at nonzero $x-a$. If a <power series> converges in a neighbourhood, choose a smaller fixed convergence radius $r$: absolute convergence bounds its tail after $N$ by $|x-a|^{N+1}\sum_{k>N}|c_k|r^{k-N-1}$ for $|x-a|\le r$. Thus a convergent <power series> is also asymptotic. The coefficients are unique, successively recovered by
$$
c_0=\lim_{x\to a}f(x),\qquad c_k=\lim_{x\to a}\frac{f(x)-\sum_{j<k}c_j(x-a)^j}{(x-a)^k}.
$$
If two expansions first differed at index $k$, division of their difference by $(x-a)^k$ would contradict this limit.
For the given <integral>, use the exact finite identity $1/(1+2xt)=\sum_{k=0}^{n-1}(-2xt)^k+(-2xt)^n/(1+2xt)$. The gamma <integral> gives
$$
\frac1{\sqrt\pi}\int_0^\infty e^{-x}x^{k-1/2}dx=\frac{\Gamma(k+1/2)}{\sqrt\pi}=\frac{1\cdot3\cdots(2k-1)}{2^k}.
$$
Consequently
$$
f(t)=\sum_{k=0}^{n-1}(-1)^k(2k-1)!!\,t^k+R_n(t),\qquad R_n(t)=\frac{(-2t)^n}{\sqrt\pi}\int_0^\infty\frac{e^{-x}x^{n-1/2}}{1+2xt}\,dx.
$$
For $t\ge0$ the denominator is at least one, so
$$
\boxed{|R_n(t)|\le(2n-1)!!\,t^n,\qquad f(t)\sim\sum_{k\ge0}(-1)^k(2k-1)!!\,t^k\quad(t\to0^+).}
$$
The empty product for $k=0$ is one. The bound is exactly the magnitude of the next omitted term. For the asymptotic definition through degree $N$, take $n=N+1$, giving an $O(t^{N+1})$ remainder. Only a finite expansion was integrated; interchanging a divergent infinite series with the <integral> would be invalid. Indeed the ratio of consecutive term magnitudes is $(2k+1)t$, so this formal series diverges at every fixed $t>0$.
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