We can indeed prove, as already suspected by მამუკა ჯიბლაძე, that $g(z)=\sum (-1)^k z^{(2k+1)^2}$ cannot be holomorphically continued past $z=1$. In fact, every point on $|z|=1$ is singular. This is a consequence of:

Theorem: Suppose that $a_n$ is bounded. Suppose further that there exists a sequence $n_j\to\infty$ such that: (1) $|a_{n_j}|\ge\delta>0$; (2) $a_{n_j-k}\to 0$ as $j\to\infty$ for every fixed $k\ge 1$. Then $\sum a_n z^n$ cannot be holomorphically continued to any open set larger than the unit disk.

(Note that $R=1$ under these assumptions.)

Of course, this applies to our power series, with $n_j=(2j+1)^2$.

A very elegant proof of the Theorem was recently given by Breuer-Simon. See reference 328 here, Theorem 1.6 of the paper.

We can indeed prove, as already suspected by მამუკა ჯიბლაძე, that $g(z)=\sum (-1)^k z^{(2k+1)^2}$ cannot be holomorphically continued past $z=1$. In fact, every point on $|z|=1$ is singular. This is a consequence of the following rather general version of the classical results on lacunary power series:

Theorem: Suppose that $a_n$ is bounded. Suppose further that there exists a sequence $n_j\to\infty$ such that: (1) $|a_{n_j}|\ge\delta>0$; (2) $a_{n_j-k}\to 0$ as $j\to\infty$ for every fixed $k\ge 1$. Then $\sum a_n z^n$ cannot be holomorphically continued to any open set larger than the unit disk.

(Note that $R=1$ under these assumptions.)

Of course, this applies to our power series, with $n_j=(2j+1)^2$.

A very elegant proof of the Theorem was recently given by Breuer-Simon. See reference 328 here, Theorem 1.6 of the paper.

We can indeed prove, as already suspected by მამუკა ჯიბლაძე, that $g(z)=\sum (-1)^k z^{(2k+1)^2}$ cannot be holomorphic continued past $z=1$. In fact, every point on $|z|=1$ is singular. This is a consequence of:

Theorem: Suppose that $a_n$ is bounded. Suppose further that there exists a sequence $n_j\to\infty$ such that: (1) $|a_{n_j}|\ge\delta>0$; (2) $a_{n_j-k}\to 0$ as $j\to\infty$ for every fixed $k\ge 1$. Then $\sum a_n z^n$ cannot be holomorphic continued to any open set larger than the unit disk.

(Note that $R=1$ under these assumptions.)

Of course, this applies to our power series, with $n_j=(2j+1)^2$.

A very elegant proof of the Theorem was recently given by Breuer-Simon. See reference 328 here, Theorem 1.6 of the paper.

We can indeed prove, as already suspected by მამუკა ჯიბლაძე, that $g(z)=\sum (-1)^k z^{(2k+1)^2}$ cannot be holomorphically continued past $z=1$. In fact, every point on $|z|=1$ is singular. This is a consequence of:

Theorem: Suppose that $a_n$ is bounded. Suppose further that there exists a sequence $n_j\to\infty$ such that: (1) $|a_{n_j}|\ge\delta>0$; (2) $a_{n_j-k}\to 0$ as $j\to\infty$ for every fixed $k\ge 1$. Then $\sum a_n z^n$ cannot be holomorphically continued to any open set larger than the unit disk.

(Note that $R=1$ under these assumptions.)

Of course, this applies to our power series, with $n_j=(2j+1)^2$.

A very elegant proof of the Theorem was recently given by Breuer-Simon. See reference 328 here, Theorem 1.6 of the paper.