## Tag Archives: non-negative coefficients

### 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 $p>k+1$.