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The following question is exercise $14$ in chapter $2$ in Stein and Shakarchi's Complex Analysis.

Suppose that $f$ is holomorphic in an open set containing the closed unit disc, except for a pole at $z_0$ on the unit circle. Show that if $$\sum_{n=0}^\infty a_nz^n$$ denotes the power series expansion $f$ in the open unit disc, then $$\lim_{n\to\infty}\frac{a_n}{a_{n+1}}=z_0.$$

A solution may be found here or here. A discussion in the comments of one of the linked questions prompted me to ask another question.

Question: Does this remain true when 'pole' is replaced by 'essential singularity'?

I strongly suspect the answer is no, but I can't find a convenient counterexample. It seems like something like $e^{\frac{1}{z-1}}$ should provide a counterexample, but the calculus involved in computing the power series coefficients is very messy. I recall there being some standard power series that people always use to show that anything can happen on the boundary of a disk of convergence, but I can't find them at the moment. Perhaps one of those works?

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Potato
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    The particular function $\exp(1/(1-z))$ is not a counterexample. The binomial series yields $\sum_{k\ge 0} (1-z)^{-k}/k! = \sum_{n\ge 0} z^n \sum_{k\ge 0}\binom{k+n-1}{n}/k! $. The sum over $k$ is $a_n={}1F_1(n;1;1)$ which grows moderately; the limit of ratio $a_n/a{n+1}$ is $1$. – ˈjuː.zɚ79365 Jun 15 '13 at 14:41
  • @ˈjuː.zɚ79365 I don't follow your computation. Could you explain a bit more? – Potato Jun 15 '13 at 14:55
  • Since it only shows an example does not work, why bother? Frankly speaking, I used WolframAlpha to sum the series for $a_n$, and I looked at the ratio $a_n/a_{n+1}$ numerically. – ˈjuː.zɚ79365 Jun 15 '13 at 15:12
  • @ˈjuː.zɚ79365 What you did is not a technique I've seen before, and I wanted to learn it. Anyway, looking at it in WA, there seem to be large groups of negative coefficients, then large groups of positive coefficients, then negative coefficients, and so on. This would mean the limit of $a_n/a_{n+1}$ doesn't exist. – Potato Jun 15 '13 at 15:54
  • I see, changing the sign of $z-1$ was a bad decision on my part. The function you have is indeed a counterexample: $\exp(1/(z-1))$ has coefficients ${}_1F_1(n;1,-1)$ which oscillate. I'll try to come up with a proof without hypergeometric machinery. – ˈjuː.zɚ79365 Jun 16 '13 at 05:01
  • @ˈjuː.zɚ79365 I'm interested in seeing the hypergeometric machinery though! – Potato Jun 16 '13 at 05:03
  • I don't really know anything apart from what's on Wikipedia. By the way, I messed up the coefficients again: it's the function $(1-z)^{-1} \exp(1/(z-1))$ that has ${}_1F_1(n+1;1;-1)$ for the coefficient of $z^n$. But I managed to put together two proofs below. – ˈjuː.zɚ79365 Jun 16 '13 at 15:41
  • One more comment: the title could be more descriptive. Something along the lines of 'asymptotics of Taylor coefficients of a function with essential singularity'. – ˈjuː.zɚ79365 Jun 16 '13 at 15:50

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Short version. The function $$g(x,z)=\frac{1}{1-z} \exp\left(-\frac{xz}{1-z}\right) \tag1$$ is the generating function for Laguerre polynomials; that is, $$g(x,z)=\sum_{n=0}^\infty L_n(x)\,z^n \tag2$$ Set $x=1$. The function $g(1,z)$ has an essential singularity at $z=1$ (it's almost the same as the function in the OP). That the ratio $L_n(1)/L_{n+1}(1)$ does not approach $1$ (or any other number) follows from the asymptotic expansion
$$L_n(1)= \frac{\sqrt{e/\pi}}{n^{1/4}} \cos(2\sqrt{n}-\pi/4)+O(n^{-3/4}) \tag3$$ which shows that $L_n(1)$ changes sign infinitely often (though increasingly slowly).

Longer but self-contained version. The function $$g(z)=\frac{1}{1-z} \exp\left(-\frac{z}{1-z}\right) \tag 4$$ satisfies $g'(z)=-z\, g(z)/(1-z)^2$, which can be written as $$g'(z)-2zg'(z)+z(zg(z))'=0 \tag5$$ Plugging $g(z)=\sum_{n=0}^\infty a_n z^n$ in (5) and extracting the coefficient of $z^n$, we find $$(n+1)a_{n+1} - 2n a_n +n a_{n-1} =0,\quad n\ge 1 \tag6 $$

Suppose $a_n$ have constant sign for $n\ge N$; without loss of generality $a_n\ge 0$ for $n\ge N$. By making $N$ larger, we can arrange $a_N>0$. Only $n\ge N$ are considered in the sequel.

Let $\Delta_n=a_{n+1}-a_n$. From (6) we have $$\Delta_{n+1} - \Delta_n = a_{n+1}-(2a_n-a_{n-1}) = -\frac{a_{n+1}}{n+1} \le 0 \tag7$$ Thus, $\Delta_n$ is a nonincreasing sequence (which makes $a_n$ a concave sequence). If some $\Delta_n$ is negative, the sequence $a_n$ will also become negative, contrary to assumption. Thus, $a_n$ is nondecreasing, hence $a_n\ge a_N$. Summing (7) over $n=N,N+1,\dots , M$ we get $$\Delta_{M+1}-\Delta_N \le -a_N \sum_{n=N}^{M}\frac{1}{n+1} \tag8$$ Since the harmonic series diverges, (6) implies $\Delta_{M+1}<0$ for large $M$, a contradiction.

Conclusion: $a_n$ changes sign infinitely often. In particular, $a_n/a_{n+1}\not\to 1$.

Remark: (6) is secretly the recurrence relation for Laguerre polynomials.

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