I am trying to show $\sum _{n=1}^{n=\infty }2^{n}\sin \frac {1} {3^{n}z}$ converges absolutely for all values of $z$ $(z=0$ excepted$)$, but does not converge uniformly near $z=0$.
I observed that $\lim _{n\rightarrow \infty }\sin \frac {1} {3^{n}z}\rightarrow 0$ and also that $\left| \sin\frac {1} {3^{n}z}\right| \leq 1$ so if we replaced all the $sin\frac {1} {3^{n}z}$ parts we end up with a geometric series and since $|2| > 1$ and in such a case it would diverges, according to wolframalpha the series converges by ratio test so $\lim _{n\rightarrow \infty }\left|\frac {2^{n+1}\sin \frac {1} {3^{n+1}z}} {2^{n}\sin \frac {1} {3^{n}z}} \right| = \lim _{n\rightarrow \infty }\left|\frac {2\sin \frac {1} {3^{n+1}z}} {\sin \frac {1} {3^{n}z}} \right|$ here i am not sure how to simplfy the sin parts further before taking the limit . I assume wolframalpha did this numerically hence the conclusion. I also noticed that at $z = 0$ the series is not defined hence it can not be uniformly convergent around z=0, but i'd really like to find out if there is an ordinary discontinuity or removable discontinuity at $z=0$.
Any help would be much appreciated.