# Test for uniform convergence of $\sum_{n=1}^{\infty} 2^{n} \sin(\frac{1}{3^{n}x})$

I have a question:

Test for the uniform convergence of the following series: $\sum_{n=1}^{\infty} 2^{n}sin(\dfrac{1}{3^{n}x})$ on $[a;+\infty)$ with $a>0$ and on $(0;+\infty)$.

For the interval $(0;+\infty)$, this is my solution :

We have $2^{n}sin(\dfrac{1}{3^{n}x})| \sim (\dfrac{2}{3})^{n}\dfrac{1}{|x|}$

We already know that $\sum_{n=1}^{\infty} (\dfrac{2}{3})^{n}\dfrac{1}{|x|}$ converges for all $x \in (0;+\infty)$

Thus $\sum_{n=1}^{\infty} 2^{n}sin(\dfrac{1}{3^{n}x})$ converges on $(0;+\infty)$.

Let $S_{n}(x)$ be the $n$th partial sum of this series. Choose $x_{n}=\dfrac{1}{3^n}$. We have $|S_{n+1}(x_n)-S_{n}(x_n)|=2^{n+1}sin\dfrac{1}{3}>sin\dfrac{1}{3}=\epsilon_0>0$

Thus by Cauchy's creterion we have this series does not convergence uniformly on $(0;+\infty)$

I wonder if we can use the similar solution for the interval $[a;+\infty)$ with $a>0$?

Let $a>0$. For the interval $[a,+\infty)$ the sequence converges uniformly. We can prove this using Weierstrass M-test.
To begin, observe that for $x\in [a,+\infty)$, $\frac{1}{3^nx}\le \frac{1}{3^na}$, which is smaller than $\pi/2$ for $n$ large enough. Hence for those large $n$, $|f_n(x)|=2^n|\sin\left(\frac{1}{3^nx}\right)|\le 2^n\sin\left(\frac{1}{3^na}\right)$.
Now, by Taylor theorem, we know $\sin(u)=u+O(u^3)$, hence, for larger $n$, (say, $n\geq n_0$ for some large $n_0$), $\sin\left(\frac{1}{3^na}\right)\sim \frac{1}{3^na}$.
Hence, for $n< n_0, |f_n(x)|\leq 2^n$ and for $n\ge n_0, |f_n(x)|\leq \left(\frac{2}{3}\right)^n\frac{1}{a}$. Since $\sum_{n<n_0}2^n+\sum_{n\geq n_0}\left(\frac{2}{3}\right)^n\frac{1}{a} <+\infty$, by Weierstrass test this sequence converges uniformly on $[a,+\infty)$.
• Thank you for your solution, Nate River. I still wonder why my solution can't be applied for the interval $[a;+\infty)$. – liverpool29 Oct 2 '16 at 13:23