Absolute and uniform convergence of $\sum _{n=1}^{n=\infty }2^{n}\sin \frac {1} {3^{n}z}$ 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.
 A: To get the absolute convergence, just use the inequality $|\sin t|\leq |t|$ for any real number $t$, which be established thanks to the fundamental theorem of analysis:$|\sin t|=|\int_0^t -\cos sds|\leq \int_0^{|t|}\cos sds\leq |t|$. 
Let $S$ a subset of $\mathbb C$ of the form $S:=\{z=0<|z|<\delta\}$, we have to show that the convergence is not uniform on $S$. To see that, note that for $n$ large enough, say larger than $n_0$, $S$ contains points of the form $3^{-n}$, so 
$\sup_{z\in S}\left|\sum_{n=N}^{+\infty}2^n\sin\frac1{3^nz}\right|\geq 2^N\sin 1,$
since the terms for $z=3^{-n}$ are non-negative if $n\geq n_0$. 
A: Here’s how you can complete the ratio test to get absolute convergence at $z\ne 0$.
Fix $z\ne 0$. Then $\left|\frac1{3^nz}\right|$ is small for sufficiently large $n$, and hence for sufficiently large $n$ we have $\sin\frac1{3^nz}\approx\frac1{3^nz}$ and
$$2^n\sin\frac1{3^nz}\approx\frac{2^n}{3^nz}=\left(\frac23\right)^n\frac1z\;.$$
In particular, you’d expect the ratio test to give a limiting ratio of $2/3$. If you want to do this a bit more carefully, make use of the known limit $$\lim_{x\to 0}\frac{x}{\sin x}=\lim_{x\to 0}\frac{\sin x}x=1$$ as follows:
$$\begin{align*}\lim_{n\to\infty}\frac{2\sin\frac1{3^{n+1}z}}{\sin\frac1{3^nz}}&=2\lim_{n\to\infty}\left(\frac{\frac1{3^nz}}{\sin\frac1{3^nz}}\cdot\frac{\sin\frac1{3^{n+1}z}}{\frac1{3^nz}}\right)\\
&=2\left(\lim_{n\to\infty}\frac{\frac1{3^nz}}{\sin\frac1{3^nz}}\right)\left(\lim_{n\to\infty}\frac{\frac13\sin\frac1{3^{n+1}z}}{\frac1{3^{n+1}z}}\right)\\
&=2\cdot 1\cdot\frac13\lim_{n\to\infty}\frac{\sin\frac1{3^{n+1}z}}{\frac1{3^{n+1}z}}\\
&=\frac23\;.
\end{align*}$$
Note that here we held $z$ fixed; to show that the convergence isn’t uniform, you’re going to have to look at what happens when $z$ is not held fixed but is merely held close to $0$, and Davide Giraudo has already covered that pretty thoroughly.
