Asymptotic power series ( $f (x) - \sum_{k = 0}^{N} c_{k} (x - a)^{k} = o (∣ x - a ∣^{N})$ )

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 \leq 0

. Every right-hand derivative is an exponential function times a polynomial in

1/ x

, tending to zero as

x ↓ 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.

Table of contents
- A divergent factorial series can approximate a convergent integral Asymptotic power series

A divergent factorial series can approximate a convergent integral ( $F (z) = \int_{0}^{\infty} e^{- t} / (1 + t / z) d t$ )

 0  0

For real

z > 0

, define

F (z) = \int_{0}^{\infty} \frac{e ^{- t}}{1 + t / z} d t .

(1)

The integrand is positive and at most

e^{- t}

, so

0 < F (z) < 1

and the improper integral converges. Substituting

s = z + t

also gives

F (z) = z e^{z} E_{1} (z)

in terms of the exponential integral; the original representation makes positivity and the large-

z

behavior transparent.

Apply the exact finite geometric series identity inside the integral, without assuming that an infinite series can be integrated term by term:

\frac{1}{1 + t / z} = \sum_{k = 0}^{N - 1} (- 1)^{k} \frac{t ^{k}}{z ^{k}} + (- 1)^{N} \frac{t ^{N} / z ^{N}}{1 + t / z} .

(2)

The Gamma integral gives

\int_{0}^{\infty} e^{- t} t^{k} d t = k!

. Therefore, with

N

retained terms,

F (z) = S_{N} (z) + R_{N} (z), S_{N} (z) = \sum_{k = 0}^{N - 1} \frac{( - 1 ) ^{k} k !}{z ^{k}},

(3)

R_{N} (z) = \frac{( - 1 ) ^{N}}{z ^{N}} \int_{0}^{\infty} \frac{e ^{- t} t ^{N}}{1 + t / z} d t, ∣ R_{N} (z) ∣ \leq \frac{N !}{z ^{N}} .

(4)

This remainder formula proves the asymptotic expansion at every fixed order:

R_{N} = O (z^{- N}) = o (z^{- N + 1})

z \to + \infty

. Its sign also gives alternating bounds: an odd

N

overestimates

F (z)

, and an even

N

underestimates it.

z = 10

, an independent evaluation of the integral gives

F (10) \approx 0.9156333394

. Three retained terms give

S_{3} (10) = 1 - 0.1 + 0.02 = 0.92, ∣ F (10) - S_{3} (10) ∣ < 0.006.

(5)

Five retained terms give

S_{5} (10) = 0.9164

with error bounded by

5! /1 0^{5} = 0.0012

. The actual errors are about

0.004367

and

0.000767

, respectively. The error bound is conservative, but established without relying on the numerical integral evaluation.

Figure 1.
Fixed-order asymptotic expansions improve as the parameter grows
. Original calculations for $F (z) = \int_{0}^{\infty} e^{- t} / (1 + t / z) d t$ . Left: the integral and its one-, three- and five-term partial sums. Right: their absolute errors, with dotted bounds $N! / z^{N}$ . The independently justified bounds apply for every displayed positive $z$ ; the asymptotic expansion concerns the limit $z \to + \infty$ . **Details**.

Nevertheless, the formal infinite series diverges at every fixed

z > 0

. The magnitudes

T_{k} = k! / z^{k}

satisfy

T_{k + 1} / T_{k} = (k + 1) / z

, so they eventually increase without bound. In particular, the terms fail the necessary zero-term condition for a convergent series. The finite remainder identity remains valid after convergence has failed; this is precisely what makes the example useful.