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For fixed $n$, define the remaindersThe hypotheses say $r_n=o({\varphi }_n)$ and $r_{n+1}=o({\varphi }_{n+1})$, while subtraction givesBecause $a_{n+1}\ne 0$,For $x$ sufficiently close to $x_0$ this denominator is therefore nonzero, andThus ${\varphi }_{n+1}=o({\varphi }_n)$ for every $n$, so $({\varphi }_n)$ is an asymptotic sequence.

Put $a=1+ϵ$ and, for $m\ge 2$, defineOne integration by parts givesStarting with $I=I_2$ and iterating $N+1$ times yieldswhereSince $0<I_{N+3}(ϵ)\le (1+ϵ{)}^{-1}$, the remainder is $O({ϵ}^{N+1})$, which is little-$o$ of the last retained term. ThereforeThe displayed terms form an asymptotic sequence because the ratio of successive magnitudes tends to zero.

The functions $({\varphi }_n)$ form an asymptotic sequence as $t\to \infty$ whenfor every $n$.

This is an asymptotic sequence. Sinceone has

For $n\ge 1$, put $r_n={\varphi }_n/{\varphi }_{n-1}$. Since $({\varphi }_n)$ is an asymptotic sequence, $r_n\to 0$. We can rewritesoTogether with ${\psi }_0={\varphi }_0$, this provesMoreover,Hence $({\psi }_n)$ is an asymptotic sequence. This is the harmonic-mean refinement of an asymptotic sequence.

An asymptotic sequence satisfiesThe expansion
$f\sim {\sum }_{n\ge 0}{a}_n{\varphi }_n$ means that for every $N\ge 0$,For $N=0$, division by ${\varphi }_0$ givesFor $n\ge 1$, the definition with $N=n$ saysDivision by ${\varphi }_n$ gives the second coefficient formula. In particular, asymptotic-expansion coefficients are unique.