# 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.

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here is an idea maybe in the ratio test case l'hospital rule can be used –  Hardy Mar 23 '12 at 21:10
What about $\sin z \sim z$? –  Pedro Tamaroff Mar 23 '12 at 21:12
I think that for $z \neq 0$ $\sum 2^n \sin \frac 1 {3^n z} < \frac{1}{z}\sum \left(\frac{2}{3}\right)^n$ –  Pedro Tamaroff Mar 23 '12 at 21:13
that's seems like a good idea too specially since z is so close to 0 –  Hardy Mar 23 '12 at 21:14

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 $n\geq n_0$.

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wow, that was an incredible answer. thank you –  Hardy Mar 23 '12 at 21:24
Is it possible to tell what kind of discontinuity exists at $z=0$ ? –  Hardy Mar 23 '12 at 21:27
If $f(z)$ is the series, what about the limit of $zf(z)$ at $0$? –  Davide Giraudo Mar 23 '12 at 21:31

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.

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thank you, your answer was very nice and educational for me. I think there are couple of typos in there but i found it very insightful. –  Hardy Mar 23 '12 at 21:39
@Hardy: I found one, where I pulled out the factor of $1/3$ but forgot to delete it inside the limit. –  Brian M. Scott Mar 23 '12 at 22:04
No worries bud i think there was a deeper beauty in the answer you posted that most would be willing to look past the minor typos. :-) –  Hardy Mar 23 '12 at 22:07