Does there exist a power series with this property?

Question

Does there exist a sequence $<a_n>$ of complex numbers such that the power series $$\sum_{n\geq0}a_n z^n$$ converges conditionally for $\mid z\mid=1$. I have never seen this phenomenon so just asking here.

Answer 1

Let $\ell(i) = \lfloor \log_2(i) \rfloor+ 1$, that is $2^{\ell(i)}$ is the least power of $2$ greater than $i$. Then it is a result of Sierpinski(1) that $$ \sum_{i=1}^\infty \frac{(-1)^{\ell(i) - 1}}{2^{\ell(i)} \ell(i)} z^i $$ converges uniformly but not absolutely everywhere on the circle $|z| = 1$ (the boundary of the disk of convergence). (This is also the last example at https://en.wikipedia.org/wiki/Radius_of_convergence#Convergence_on_the_boundary .) Note that the function of $\ell$ is to produce repetitions of coefficients for powers of $z$ lying between consecutive powers of $2$.

For $|z| = 1$, \begin{align*} \sum_{i=1}^\infty &\left|\frac{(-1)^{\ell(i) - 1}}{2^{\ell(i)} \ell(i)} z^i \right| \\ &= \frac{1}{2} + \frac{1}{8} + \frac{1}{8} + \frac{1}{24}+ \frac{1}{24}+ \frac{1}{24}+ \frac{1}{24} + {\cdots} \\ &= \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots \text{,} \end{align*} where we group identical terms to obtain half the harmonic series, showing failure of absolute convergence at every point on the circle $|z| = 1$.

Sierpinski passes through a sequence of estimates to show that the magnitude of the difference between the $q^\text{th}$ and $p^\text{th}$ partial sums ($p<q$) decreases like $1/p$, establishing uniform convergence on the entire circle $|z| = 1$, hence conditional convergence on that circle.

(1) Sierpiński, Wacław (1918), "O szeregu potęgowym który jest zbieżny na całem swem kole zbieżności jednostajnie ale nie bezwzględnie", Prace matematyka-fizyka, 29, pp. 263–266.