Power series infinity at every point of boundary Is there an example of a power series $f(z)=\sum_{k=0}^\infty a_kz^k$ with radius of convergence $0<R<\infty$ so that $\sum_{k=0}^\infty a_kw^k=\infty$ for all $w$ with $|w|=R$ 
Thank you kindly.
 A: Consider a power series $f(z)=\sum_{n=0}^\infty a_n z^n$ with radius of convergence $0<R<\infty$.
Assume $\sum_{n=0}^\infty a_n z^n=\infty$ for all $z$ with $|z|=R$.
There are at most finitely many zeroes $z_1,\ldots,z_m$ of $f$ in the open disk $D=\{z\colon|z|<R\}$ so that for a suitable polynomial $p$  we obtain a function $g\colon D\to\mathbb C$, $g(z)=\frac{p(z)}{f(z)}$ that is holomorphic on $D$ and $g(z)\to0$ (why?) as $|z|\to 1$, hence is bounded. Using the reflection principle we get an entire bounded function, hence $g$ is constant.
But then the power series of $f$ is better-behaved on $\partial D$, contradiction. 
A: No there is not. In fact, there is no example of such power series $\sum_n a_n z^n$ such that $\sum_n a_n w^n = \infty$ for all $w$ in a set of positive measure in $\partial D$, where $D=\{|z|<R\}$. Indeed, suppose there exists such a power series $f$. By Abel's Theorem, we deduce that $f(z)$ has non-tangential boundary values $\infty$ on a set of positive measure in $\partial D$. This means that $1/f$ is a meromorphic function in $D$ with non-tangential boundary values $0$ on a set of positive measure in $\partial D$, and so $1/f$ is identically zero in $D$, by the Luzin-Privalov Theorem. So $f \equiv \infty$ in $D$, a contradiction.
A: For the series $\sum\limits_{n=1}^{\infty}{z^n}$ radius  of convergence  $R=1$, but it diverges $\forall{z}\colon \;\;|z|=1.$
A: Take the series $\sum_k z^k$. its radius of convergence $R=1$ and for $z=e^{i\theta}$
the series $\sum_n e^{in\theta}$ diverges since the term $e^{in\theta}$ does not converge to $0$
A: The series $$\sum_{k=1}^\infty k \cdot z^k $$ has convergence radius $1$ and for $z=1$ the absolut value of the series goes to infty. 
