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An asymptotic expansion describes a function near a specified limit by successively smaller scales. Each fixed finite truncation has a remainder smaller than its last retained scale. The central question is how the approximation improves as the parameter approaches the limit; convergence of the entire infinite series at a fixed parameter is a separate question. For the inverse-power scale $1,z^{-1},z^{-2},\dots$,means that, for every fixed nonnegative integer $m$,The little o notation makes the accuracy requirement precise. With expansions to all orders, this is also equivalent to an $O(z^{-m-1})$ remainder after the $m$th term; the implied bound may depend on $m$.^[ref]

For a quick convergent example, the finite geometric series identity givesFor $z>1$, both the asymptotic expansion as $z\to +\infty$ and the ordinary convergent geometric series describe this function. The exact remainder displays the required extra inverse power.

The more surprising example under a divergent factorial series can approximate a convergent integral has a convergent integral but a divergent series at every fixed positive $z$:A few terms can estimate $F(10)$ well, even though adding sufficiently many terms eventually destroys the approximation. Optimal truncation explains where to stop.

This makes asymptotic expansions useful when an exact expression is a complicated integral or differential equation solution, but a parameter is very large or very small. The leading term reveals the scale; corrections reveal which effects matter next. The asymptotic sequence must be specified: integer powers, fractional powers, logarithms, and exponential functions can encode different kinds of behavior. A good finite approximation includes a remainder estimate or an independently checked range of use.

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 sequence orders the scales against which successive errors are compared. On a specified domain approaching $x_0$, suppose each ${\varphi }_n$ is nonzero sufficiently close to the limit andEquivalently, ${\varphi }_{n+1}=o({\varphi }_n)$ in little o notation. The notationmeans that for every $N\ge 0$,Each $N$ stays fixed while the parameter tends to its limit.^[ref]

As $z\to +\infty$, $1,z^{-1},z^{-2},\dots$ is an asymptotic sequence, because each adjacent ratio is $1/z$. Fractional scales $1,z^{-1/2},z^{-1},\dots$ work too, with ratio $z^{-1/2}$. By contrast, $1,1+z^{-1},z^{-1},\dots$ fails at its first pair: the second scale is not smaller than the first. Ordering by written position alone does not create a hierarchy of sizes.

Successive division of the remainder by ${\varphi }_N$ uniquely recovers every coefficient:If two proposed expansions first differ at order $N$, subtracting them and dividing by ${\varphi }_N$ would give a nonzero constant tending to zero, a contradiction. This is uniqueness of the coefficients in a fixed scale, not uniqueness of the represented function.

For the factorial integral example, the first three limiting responses are $\lim F(z)=1$, $\lim z(F(z)-1)=-1$, and $\lim z^2(F(z)-1+z^{-1})=2$. They extract exactly the coefficients $1,-1,2!$. Choosing an asymptotic sequence therefore turns “the next small correction” into a precise and reusable operation.

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.

For every fixed truncation index $N$, the remainder must be smaller than the last retained term:This is an asymptotic expansion relative to the sequence in part (i). The number of retained terms is fixed before taking the limit. The associated series need not converge for any nonzero parameter; its role is to give successively smaller truncation errors. If $a_n$ is an asymptotic sequence, the same condition can be stated as a remainder $O(a_{N+1})$ when the next expansion condition is available, but the displayed little o notation formulation is the defining one.

For each fixed $n$, the next term must be negligible relative to its predecessor:The ratios are understood where their denominators are nonzero near the limiting point, or equivalently through the little o notation definition. For example $1,ϵ,{ϵ}^2,\dots$ form an asymptotic sequence.

For an asymptotic sequence $a_{n+1}(ϵ)=o(a_n(ϵ))$, the asymptotic expansion condition is that every fixed truncation has remainder smaller than its last retained term:This limit is taken as $ϵ\to 0$ in the specified direction or sector. The number $N$ stays fixed while taking that limit. The infinite series need not converge for any nonzero parameter.