Uniform convergence

1. May 11, 2005

twoflower

Hi,

I don't know how to analyse uniform convergence/local uniform convergence for this series of functions:

$$\sum_{n=1}^{\infty}2^{n}\sin \frac{1}{3^{n}x}$$

Then
$$f_{n}^{'} = -\left(\frac{2}{3}\right)^{n}\frac{\cos \frac{1}{3^{n}x}}{x^2}$$
$$f_{n}^{'} = 0 \Leftrightarrow x = \frac{1}{3^{n}\left(\frac{\pi}{2} + k\pi\right)}$$

I tried to prove the convergence using the theorem about change of sum and derivation - if I showed that
$$\sum_{n=1}^{\infty} f_{n}^{'} \rightrightarrows^{loc} on (a,b)$$

I could tell that also
$$\sum_{n=1}^{\infty} f_{n} \rightrightarrows^{loc} on (a,b)$$

Maybe I could use Dirichlet/Abel to show the convergence, but I'm not sure I have met the preconditions. Anyway, if I wanted to show directly convergence of the original series using the most straightforward criterion - Weierstrass, could I write it this way: ?

Maximum for $f_{n}$ is in $x = \frac{1}{3^{n}\left(\frac{\pi}{2} + 2k\pi\right)}$. Then
$$S_{n} := \sup_{x \in (0,\infty)} \left|2^{n}\sin \frac{1}{3^{n}x}\right| = 2^{n}$$

But
$$\sum_{n=1}^{\infty} S_{n} = \infty$$
so I haven't shown the convergence. Anyway, as we can see, x at which our function has the maximum is moving to zero as n grows. So, what if I chose another interval for x, let's say $x \in (\epsilon, \infty), \epsilon > 0$? Then

$$\exists n_0 \in \mathbb{N} \mbox{ such that } \forall n \ge n_0\ \ \ \ \ \ \ \frac{1}{3^{n}\left(\frac{\pi}{2} + 2k\pi}\right)} < \epsilon$$

I feel this is maybe good idea, but don't know how to pull it into finish or how to use this idea...