Let $\displaystyle f(z)=\sum_{n=0}^\infty z^{2^n}$ for $|z|<1$
Show that $f$ cannot be continued analytically past the unit disc.
I didn't figure out about a solution I've found.
Show that $f(z)=\sum_{n=0}^{\infty}z^{2^n}$ can't be analytically continued past the unit disk.
The solution :
Yes, your "I understand from the hint" paragraph is good (I especially liked "plowing ahead"). Working with the sum, I wouldn't right off the bat start taking limits. What we can say is that for $0\le r <1,$
$$\tag 1\left| \sum_{n=0}^\infty r^{2^n}e^{ i2\pi p 2^{n-k}}\right| \ge \left |\sum_{n=k+1}^\infty r^{2^n}e^{ i2\pi p2^{n-k}} \right| - \left| \sum_{n=0}^k r^{2^n}e^{ \frac{i2\pi p}{2^{k-n}}}\right|.$$
Now the $n$th summand in first sum on the right is simply $r^{2^n}.$ (That's why those weird points on the boundary were chosen.) In the second sum, move the absolute values inside the sum. This shows $(1)$ is at least
$$\sum_{n=k+1}^\infty r^{2^n} - \sum_{n=0}^k r^{2^n} \ge \sum_{n=k+1}^\infty r^{2^n} - (k+1).$$
Thus we're done if we show the first sum on the right has limit $\infty.$ One way to do this is let $N\in \mathbb N, N>k+1 .$ Then
$$\sum_{n=k+1}^\infty r^{2^n} > \sum_{n=k+1}^N r^{2^n}.$$
This implies
$$\tag 2 \lim_{r\to 1^-} \sum_{n=k+1}^\infty r^{2^n} \ge N-k.$$ Since $N$ is arbitrarily large, the left side of $(2)$ is $\infty,$ and we're done.
I didn't figure out from step 1 to step 2.
This : Now the n th summand in first sum on the right is simply r2n. (That's why those weird points on the boundary were chosen.) In the second sum, move the absolute values inside the sum. This shows (1) is at least.
Why did he get rid of $e^{ i2\pi p 2^{n-k}}$ and $e^{ \frac{i2\pi p}{2^{k-n}}}$. Appreciate any help.
