Asymptotic expansion of $\ln (1-\ln(\varepsilon))$

Question

Question:

Find an asymptotic expansion for $\ln (1-\ln(\varepsilon))$ as $\varepsilon \rightarrow 0$.

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Attempt:

I don't even understand what exactly I am supposed to do.

I am not given an asymptotic sequence for which to solve the coefficients.

$\ln(1-\ln(\varepsilon))$ obviously blows up as $\varepsilon \rightarrow 0$, so I'm not even sure what an appropriate asymptotic sequence should be (the usual power series is obviously not going to work).

Any hints?

Answer 1

You have

$$\begin{aligned} f(\epsilon)=\ln (1-\ln(\varepsilon)) &= \ln \left(\frac{-\ln(\varepsilon)}{-\ln(\varepsilon)}(1-\ln(\varepsilon))\right) = \ln(-\ln(\varepsilon)) + \ln \left(1 - \frac{1}{\ln \varepsilon} \right)\\ &= \ln(-\ln(\varepsilon)) - \frac{1}{\ln \varepsilon} - \frac{1}{2\ln^2 \varepsilon}- \frac{1}{3\ln^3 \varepsilon} -\dots \end{aligned}$$

which is the desired asymptotic expansion.