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I need to show that $\phi_n(z)=\ln(1+z^n)$ as $z \rightarrow 0$ is an asymptotic sequence, i.e. to show that $\lim_{z\rightarrow 0}\frac{\phi_{n+1}(z)}{\phi_n(z)}=0.$

Is it sufficient for me to say that as $z\rightarrow 0$, $\frac{\phi_{n+1}}{\phi_n}=\frac{\ln(1+z^{n+1})}{\ln(1+z^n)} \rightarrow 0?$ Because $z^{n+1}$ and $z^n \rightarrow 0$ as $z\rightarrow 0$, and $\ln(1)=0?$

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No, the $0/0$ form is indeterminate so you can't simply plug the limits into the logarithm. Consider the limits of things like $z^2/z$, $z/z$, and $z/z^2$: in each case numerator and denominator $\to0$, but the limits of them are $0$, $1$ and $\infty$ respectively. This looks like an excellect application of l'Hospital's.

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    Than$k$s anon, I and shall work on it, appreciated!2012-02-23
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Just use the formula $\log (1+w) = w + \mathcal{O}(w^2)$,as $\mathbb C \ni w \to 0$.