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Let $f$ be a function holomorphic in an open set containing the closed unit disc $D(0,1)$, except at the point $z_0$ with $|z_0|=1$, where $f$ has an isolated singularity. If $a_n$ are the coefficients of the Taylor expansion of $f$ centered at the origin, show that

$$\displaystyle\lim_{n \to\infty}\frac{a_n}{a_{n+1}}=z_0$$

I considered first the case: $z_0$=simple pole. Then i can write the Laurent series for $f$ at $z_0$

$$f(z)=\frac{c}{z-z_0}+c_0+c_1(z-z_0)+\ldots$$ I put


Then $g$ is holomorphic in an open set containing the closed unit disc $\overline{D(0,1)}$ and we have $$g(z)=\sum a_n z^n+\frac{c}{z_0}\sum(\frac{z}{z_0})^n=\sum(a_n+\frac{c}{z_0^{n+1}})z^n$$ The power series of $g$ has radius of convergence $>1$, hence i can substitute 1 in RHS of last equation, and i get a numeric convergent series. The necessary condition for convergence says that its general term has to be infinitesimal $$\displaystyle\lim_{n\to\infty}(a_n+\frac{c}{z_0^{n+1}})=0$$ from which follows $\frac{a_n}{a_{n+1}}\rightarrow z_0$

the question is:how can i adapt this argument to deal with cases where z_0 is not a simple pole?

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marked as duplicate by Potato, Aaron Maroja, Jeel Shah, Jonas Meyer, Did Apr 22 at 20:44

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It's certainly not true if $z_0$ is a removable singularity. –  Robert Israel Nov 23 '12 at 19:32
It is true for a pole, but I'm not convinced it's true for an essential singularity. –  Robert Israel Nov 23 '12 at 19:39
why not for removable singularities? –  Federica Maggioni Nov 23 '12 at 21:52
Because the Taylor series would be the same if you removed the singularity by setting $f(z_0) = \lim_{z \to z_0} f(z)$. –  Robert Israel Nov 25 '12 at 2:49
let's see if i've understood: i take a function with removable singularity of absolute value 1, i remove the singularity by substituting the limit as you said, then i get an holomorphic function in an open set containing the closed unit disc centered at the origin, hence the radius of convergence of its Taylor expansion (equal to the original one for identity principle) is greater than 1, thus $\frac{a_n}{a_{n+1}}$ tends to a complex number whose modulus is $>1$, hence cannot be $z_0$ –  Federica Maggioni Nov 25 '12 at 8:20

1 Answer 1

up vote 1 down vote accepted

For convenience, take $z_0 = 1$.

Let $\Gamma$ be the positively oriented circle $|z|=1+r$, where $r > 0$ and $f$ is analytic in $\{z: |z| \le 1+s\} \backslash \{1\}$ for some $s > r$. Let $$ c_n = \frac{1}{2\pi i} \oint_\Gamma \frac{f(z)}{z^{n+1}}\ dz,\ b_n = \text{Res}(f(z)/z^{n+1}; z=1)$$

By the residue theorem, $a_n = c_n - b_n$. Now $|c_n| < M (1+r)^{-n-1}$ for some constant $M$.

Suppose the Laurent series of $f(z)$ in $\{0 < |z - 1| < r\}$ is $\sum_{k=-\infty}^\infty d_k (z - 1)^k$. Since $z^{-n-1} = \sum_{j=0}^\infty {{n+j} \choose n} (1-z)^j$, we get $$ b_n = \sum_{j=0}^\infty {{n+j} \choose n} (-1)^j d_{-j-1}$$

If $1$ is a pole of order $m$, this is a finite sum and the dominant term is for $j=m-1$, namely $\displaystyle {{n+m-1} \choose n} (-1)^{m-1} d_{-m} = \frac{n^{m-1}}{(m-1)!} d_{-m} + O(n^{m-2})$ and we get $$\frac{a_n}{a_{n+1}} = \frac{n^{m-1} d_{-m} + O(n^{m-2})}{(n+1)^{m-1} d_{-m} + O(n^{m-2})} = 1 + O(1/n)$$

But it's not at all clear to me what will happen in the case of an essential singularity. It seems entirely possible, for example, that some cancellations will make infinitely many $a_n = 0$.

EDIT: Consider $f(z) = \exp(1/(z-1))$. This has its only singularity at $z=1$. I claim that there are infinitely many positive $a_n$ and infinitely many negative $a_n$, which implies that $a_n/(a_{n+1}) \le 0$ infinitely often.

Since each derivative $f^{(n)}(z)$ is a rational function of $z$ (whose only pole is at $1$) times $f(z)$, we have $f^{(n)}(z) \to 0$ as $z \to 1-$. But if only finitely many coefficients $a_j$ were positive or only finitely many were negative, all the coefficients of $f^{(n)}(z)$ would have the same sign for sufficiently large $n$, and then $\lim_{z \to 1-} f^{(n)}(z)$ could not be $0$.

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