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This is a theorem of Flajolet and Odlyzko (I think): Let $f(z)$ be a function analytic in a domain $$D = \{z : |z| \leq s_1, |\text{Arg}(z-s)| > \frac{\pi}{2} - \eta \},$$ where $s, s_1 > s,$ and $\eta$ are three positive real numbers. Assume that, with $ \sigma(u) = u^\alpha \log^\beta u$ and $\alpha \notin \{0, -1, -2, \dots \}$, we have $$ f(z) \sim \sigma \left( \frac{1}{1-z/s} \right) \qquad \text{ as } z \rightarrow s \text{ in } D.$$ Then, the Taylor coefficients of $f(z)$ satisfy $$[z^n]f(z) \sim s^{-n} \frac{\sigma(n)}{n \Gamma(\alpha)}.$$

My question is whether, with the same premise, except that $\alpha = 0, \beta \in \{1, 2, 3, \dots\}$, we have $$[z^n]f(z) \sim \beta s^{-n} n^{-1} \log^{\beta-1} n.$$

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up vote 1 down vote accepted

No. For a counterexample, take $i=\sqrt{-1}$, $$f(z)= C(1-z)^i + \log [(1-z)^{-1}],$$ and let the domain $D$ be as above with the parameters $s=1$, $s_1=2$, $\eta=\pi/4$. In $D$, we have $$|(1-z)^i| = |e^{i \log (1-z)}| = e^{-\arg (1-z)},$$ which is bounded. Therefore, as $z\to 1$ in $D$, $f(z)\sim \log [(1-z)^{-1}]$. However, by Theorem VI.2 (p. 385, [Flajolet and Sedgewick 2009]), $$[z^n]f(z)=\frac{1}{n}+C [z^n] (1-z)^i =\frac{1}{n}(1+C\frac{e^{-i\log n}}{\Gamma(-i)}+O(\frac{1}{\log n})),$$ so $[z^n]f(z)$ is not asymptotic to $1/n$.

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