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I have a power series whose coefficients are all positive integers and whose radius of convergence $r$ is $<1$ and I wish to prove that it has a pole at $r$, or at least an infinite radial limit. Is there a general result that could help in this situation or should I look for an ad-hoc proof?

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I think one can give a simpler proof of this assertion in your particular case. –  Pedro Tamaroff Jul 17 '13 at 19:09
5  
Not true, the coefficients of power series of $\frac{1}{\sqrt{1-4z}} = \sum_{k=0}^{\infty}\binom{2k}{k} z^k$ are all positive integers, its radius of convergence $r = \frac14 < 1$ and yet it has a branch cut instead of pole at $r$. –  achille hui Jul 17 '13 at 19:36

2 Answers 2

up vote 4 down vote accepted

I can give you the following proof of what is an extension of Abel's Theorem.

THM Suppose $\alpha_n$ is a sequence of real numbers such that $\sum_{n\geqslant 0}\alpha_n$ diverges to $+\infty$, and such that its powerseries converges for $|x|<1$. Then $$\lim_{x\to 1^{-}}\sum_{n\geqslant 0}\alpha_nx^n=+\infty$$

P Let $M>0$ be given. Note that $$\frac{1}{1-x}f(x)=\sum_{n\geqslant 0}\sum_{k=0}^n\alpha_k x^n$$

By hypothesis this is true for $|x|<1$. Also, there exists $N>0$ such that $n>N$ implies $\sum_{k=1}^n\alpha_k>M$. Then we have that $$\displaylines{   \frac{1}{{1 - x}}f(x) = \sum\limits_{n \geqslant 0} {\sum\limits_{k = 0}^n {{\alpha _k}} } {x^n} \cr     = \sum\limits_{n = 0}^N {\sum\limits_{k = 0}^n {{\alpha _k}} {x^n}}  + \sum\limits_{n > N} {\sum\limits_{k = 0}^n {{\alpha _k}} } {x^n} \cr     > \sum\limits_{n = 0}^N {\sum\limits_{k = 0}^n {{\alpha _k}} {x^n}}  + M\sum\limits_{n > N} {{x^n}}  \cr     > M{x^{N + 1}}\frac{1}{{1 - x}} \cr} $$ It follows that $f(x) > M{x^{N + 1}}$ so $$\mathop {\lim \inf }\limits_{x \to {1^ - }} f(x) \geqslant M$$ for each $M>0$, whence it must be the case

$$\mathop {\lim }\limits_{x \to {1^ - }} f(x) = + \infty $$

ADD Note the last inequality follows from the fact we can assume the partial sums are all positive.

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That would prove that the radial limit infinity as $z \to 1^-$. Does that necessarily imply that there is a pole at that point? –  Old John Jul 17 '13 at 19:26
    
@JohnWordsworth Hmm... I misread the question then, because as you can see, I am assuming $z$ is real. So this proves there is a "real pole" but not more. –  Pedro Tamaroff Jul 17 '13 at 19:29
    
@JohnWordsworth Also: drop by the chat! Come on now! –  Pedro Tamaroff Jul 17 '13 at 19:32
    
Not sure I have time to drop into chat right now! I might have read more into the question than was intended - I assumed that "radius of convergence" meant we were in the complex plane, and then we get into things like Fatou's theorem: en.wikipedia.org/wiki/Fatou's_theorem –  Old John Jul 17 '13 at 19:35
    
@anon The inequality follows from the fact the left hand side contains at least the amount of positive elements the right hand side does, that is all. –  Pedro Tamaroff Jul 17 '13 at 19:45

You are looking for Pringsheim's theorem. I'm citing from Wikipedia's entry on Alfred Pringsheim: "One of Pringsheim's theorems, according to Hadamard [1] earlier proved by E. Borel, states [2] that a power series with positive coefficients and radius of convergence equal to 1 has necessarily a singularity at the point 1.". Of course you also get the same statement when the radius of convergence is $r$ by scaling the coefficients.

EDIT: Precisely if $f$ has positive coefficients and radius of convergence $r$ then $g(z) = f(z r)$ has radius of convergence $1$ and positive coefficients, therefore by Pringsheim's theorem $g$ has a singularity at $1$, hence $f$ has a singularity at $r$.

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Having a singularity doesn't imply the singularity is a pole. eg. $\frac{1}{\sqrt{1-z}}$. –  achille hui Jul 17 '13 at 19:31
    
Okay, fine, I thought of "singularity" when I wrote "pole". Either way Pringsheim's theorem is the theorem the OP is looking for, since without further information nothing more can be really said about the nature of the singularity at $z = r$. –  blabler Jul 18 '13 at 17:18

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