Tag Archives: nonextentable

Pringsheim Theorem

\mathbf{Thm(Pringsheim):} Suppose f(z)=\sum\limits_{n=0}^{\infty}a_nz^n has convergence radius 1. If all a_n\geq 0, then z=1 is a singular point of f.

\mathbf{Proof:} Expand f as Taylor series at z=\frac 12

\displaystyle f(z)=\sum \limits_{k=0}^{\infty}(z-\frac 12)^k\sum\limits_{n=k}^{\infty}C_n^ka_n(\frac 12)^{n-k}

If f has holomorphic extension on z=1, then the above series must converge for some z=1+\epsilon, \epsilon>0, small enough,

\displaystyle f(1+\epsilon)=\sum \limits_{k=0}^{\infty}(\frac 12+\epsilon)^k\sum\limits_{n=k}^{\infty}C_n^ka_n(\frac 12)^{n-k}

Since the double series has positive terms, we can interchange the order of two summations without loss of convergence. Hence the series,

\displaystyle\sum \limits_{n=0}^{\infty}a_n\sum\limits_{k=0}^{n}C_n^ka_n(\frac 12+\epsilon)^k(\frac 12)^{n-k}=\sum \limits_{n=0}^{\infty}a_n(1+\epsilon)^n

must converge. This contradicts the fact that f has convergence radius 1. \square
\mathbf{Remark:}

  • This theorem is posted by Alfred Pringsheim. But his proof is not right. For correct proof, refer to Analytic function Volume 1 by Einar Hille.
  • This theorem concludes there is at least one singular point fo the power series on its circle of convergence.
  • Consider f_p(z)=\sum\limits_{n=0}^{\infty}\frac{1}{n^p}z^n, p is real, which has convergence radius 1. This theorem applies and shows that z=1 is a singular point of f_p(z). actually, it is the only singularity on |z|=1. In this case, f_p(z)\to \infty, as z\to 1 when p\leq 1 but not for p>1. f^{(k)}(z)\to \infty as z\to 1 when p\leq k+1 but not for p>k+1.
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