Asymptotic series as x approaches infinity

[rg2004]

Homework Statement

Find the first 3 (non-zero) terms of the asymptotic series as x→∞ for ln(e^x+1)

Homework Equations

ln(1+ε) ~ ε-ε²/2+ε³/3-O(ε⁴) (x→0)

The Attempt at a Solution

I found

x+ln(1+e^-x) (x→∞) = ln(e^x+1) (x→∞)​

ε=e^-x​

which allows me to use the Maclaurin series above (since now the ε→0 as x→∞) yielding
x+[e^-x-e^-2x/2+e^-3x/3+O(e^-4x)]

My question
Is there some expansion I should be using for e^-x (x→∞) or some rearrangement of the e^x (x→0) that I can use as replacement for e^-x to find the next two terms of the series

Thanks
Clay

 

[susskind_leon]

You could set x'=1/x to express ln(1+e^(-x)) in terms of x' for x'->0

 

[Mute]

My question
Is there some expansion I should be using for e^-x (x→∞) or some rearrangement of the e^x (x→0) that I can use as replacement for e^-x to find the next two terms of the series?

No. The asymptotic expansion of exp(-x) for large x is just exp(-x). The taylor series expansion is valid only for small x, so there's nothing else you can do. If you're looking for an expansion that looks like

$$\ln(1+e^x) \sim f(x)\left( 1 + \frac{a_1}{x} + \frac{a_2}{x^2} + \mathcal O \left(\frac{1}{x^3}\right) \right),$$

then I don't think that's possible. (Mathematica, seems to agree, suggesting there is no expansion of the function around x = infinity). I'd say the best you could do is define y = e^x and then modify the asymptotic series you already derived that's expressed as ln(y) + inverse powers of e^x = y.