Let $a_k $ be complex numbers such that the radius of convergence of $\sum a_k z^k$ is $1$.
Suppose $|\sum_{k=0}^{N} a_k |\longrightarrow +\infty $ as $N\to +\infty$.
Then, is it true that $$\lim_{z\to 1,\, z\in \mathbb R}\left| \sum_{k=0}^{+\infty} a_k z^k \right|=+\infty\,? $$
Here, the question was answered for real $a_k$.
Thank you in advance.
We can have
$$\Biggl\lvert \sum_{k = 0}^{N} a_k\Biggr\rvert \to +\infty$$
even if
$$f(z) = \sum_{k = 0}^{\infty} a_kz^k$$
has an analytic continuation to a neighbourhood of $1$, which clearly implies that $f(z)$ remains bounded as $z \to 1$. One of the simplest examples is $f(z) = \frac{1}{(1+z)^2}$, which has the Maclaurin series
$$\sum_{k = 0}^{\infty} (-1)^k(k+1)z^k\,.$$
The partial sums of the coefficients are
$$\sum_{k = 0}^{N} (-1)^k(k+1) = (-1)^{N}\biggl\lfloor \frac{N+2}{2}\biggr\rfloor\,.$$
If however
$$\sum_{k = 0}^N a_k$$
tends to $\infty$ in a specific direction, say
$$\lim_{N \to \infty} \operatorname{Re} e^{-i\varphi}\sum_{k = 0}^{N} a_k = +\infty\,,$$
then it follows that
$$\lim_{\substack{z \to 1 \\ z \in (0,1)}} \operatorname{Re} e^{-i\varphi}f(z) = +\infty\,.$$
This is an easy consequence of the real case of course.