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An asymptotic power series uses powers as its asymptotic sequence. Near a finite point $a$, the condition in the title applies along the specified domain for every fixed $N$; at infinity, inverse powers are convenient. It need not be a convergent power series. Its coefficients record the successive limiting responses, as the integral example below demonstrates.

The finite-order Taylor theorem supplies an asymptotic power series for a smooth function near a point, but smoothness does not imply that its Taylor series converges to the function. Define $g(x)=e^{-1/x}$ for $x>0$ and $g(x)=0$ for $x\le 0$. Every right-hand derivative is an exponential function times a polynomial in $1/x$, tending to zero as $x\downarrow 0$. Consequently $g$ is a smooth function with every derivative zero at the origin. Its all-zero Taylor series misses $g(x)>0$ to the right. Zero and $g$ also have the same all-zero asymptotic power series on that side, despite being different functions. An asymptotic expansion can determine all its own coefficients without determining the represented function uniquely.

An asymptotic power series about $a$ means that for every fixed $N\ge 0$,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 byIf 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 givesConsequentlyFor $t\ge 0$ the denominator is at least one, soThe 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 $t>0$.