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I'm given the following problem. Let $f(x)=\sum c_nz^n$ in the unit disk, and furthermore $c_n\in \mathbb{R}, \ c_n >0$. Then the point $z_0=1$ is singular.

I don't seem to understand this, because if we let $c_n=2^{-n}$, then the sum $\sum 2^{-n}$ converges, so that would be a counterexample.

What am I doing wrong here?

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What are the bounds of summation? And are you sure you're not overlooking some condition on $c_n$? I agree that $\sum_{n=0}^{\infty} 2^{-n}$ converges. – Dimitrije Kostic Nov 25 '11 at 18:43
up vote 3 down vote accepted

I'm sure that you want to assume, not just that the radius of convergence is at least 1, but that the radius of convergence is exactly 1. In this case the theorem is apparently due to Vivanti (1893) and Pringsheim (1894).

I believe the proof by contradiction goes as follows: if $f$ is analytic at $z=1$, then it is analytic in a small disk centered at $z=1$. It follows that the power series of $f$ at $z=1/2$ converges in a disk of radius slightly greater than $1/2$. Choose a point $z_1>1$ at which this last series converges, and then rewrite the coefficients of that series (which are derivatives of $f$ at $z=1/2$) in terms of the original series $\sum c_n z^n$ at $z=0$. When you rearrange the resulting double sum (this is where the nonnegativity hypothesis is used!), you conclude that the original series converges at $z_1$, which contradicts the assumption that the radius of convergence is exactly 1.

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