Proving that the series $\sum\limits_{n=0}^{\infty} 2^n \sin (\frac{1}{3^nx})$ does not converge uniformly on $(0,\infty)$

Question

Show that the series $$\sum_{n=0}^{\infty} 2^n \sin (\frac{1}{3^nx})$$ does not converge uniformly on $(0,\infty)$

Solution:

My question is what is the technique being used in the solution to show that the series does not converge uniformly? Can anyone explain it with reference to the relevant definition/theorem?

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What I don't understand is that how showing that $$\sum_{n=N}^{\infty} 2^n \sin (\frac{1}{3^nx})$$ cannot be made arbitrarily small proves the result. This answer seems more like its showing that the sequence of partial sums is unbounded. I need clarification.

Answer 1

what is the technique being used in the solution to show that the series does not converge uniformly

Go back to the definition. I will explain it abstractly here.

Consider a sequence of functions $f_n:E\to{\bf R}$, where $E$ is some subset of ${\bf R}$. There are two basic modes of convergence of the sequence $\{f_n\}$.

• If for each $x\in E$, $f_n(x)\to f(x)$ as $n\to\infty$, we say that $f_n$ converges to $f$ pointwise. In terms of the $\epsilon$-$N$ language, $$ \forall x\in E\ \forall \epsilon>0\ \exists N\in{\bf N}\ \forall n\geq N\ \ |f_n(x)-f(x)|<\epsilon\tag{1} $$
• We say $f_n$ converges to $f$ uniformly, if the following is true: $$ \forall \epsilon>0\ \exists N\in{\bf N}\ \forall n\geq N\ \forall x\in E\ \ |f_n(x)-f(x)|<\epsilon\tag{2} $$

Note that (2) is a stronger condition than (1).

Remember that a series is nothing but the limit of a partial sum. Suppose $g_n:(0,\infty)\to{\bf R}$, then $\sum_{n=0}^\infty g_n(x)$ converges uniformly means, the sequence of functions $f_k$, defined with $$ f_k(x):=\sum_{n=0}^k g_n(x), $$ converges uniformly to some function $g:(0,\infty)\to{\bf R}$.

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Now, note that $f_k$ does not converge uniformly on $(0,\infty)$ means (2) is not true, namely the following holds: $$ \exists \epsilon>0\ \forall N\in{\bf N}\ \exists n\geq N\ \exists x\in E\ \ |f_n(x)-f(x)|\geq \epsilon\tag{3} $$

It is an instructive exercise to check how your "solution" matches (3).

Answer 2

By definition: a series $\sum_{ n=0}^{\infty} a_n(x)$ converges uniformly iff: $$ \lim_{N\to \infty}\sup_{x\in\Bbb R} \left|\sum_{ n=N}^{\infty} a_n(x)\right| =0$$

see here: Explain $\lim_{N\to \infty}\sup_{x\in\Bbb R} \left|\sum_{ n=N}^{\infty} a_n(x)\right| =0$

They essentially used the following well known inequality

For all $x\in[0,\frac\pi2]$ We have, $$\frac{2}{\pi}x\le \sin x\le x$$

As follows: for very large $N\in\Bbb N,$ Taking $ \color{blue}{x =\frac{2}{3^N\pi}}$ and using the geometric sum we obtain:

$$\infty =\sup_{x\in(0,\infty)} \left|\sum_{ n=N}^{\infty} 2^n \sin (\frac{1}{3^nx}) \right| \overset{\color{blue}{x =\frac{2}{3^N\pi}} }{\ge} \sum_{ n=N}^{\infty} 2^n \sin (\frac{3^N\pi}{3^n2})\\\ge \sum_{ n=N}^{\infty} 3^N \left(\frac{2}{3}\right)^n= 3.2^N.$$

That is $$\infty = \lim_{N\to \infty}\sup_{x\in(0,\infty)} \left|\sum_{ n=N}^{\infty} 2^n \sin (\frac{1}{3^nx}) \right| \ge \lim_{N\to \infty}2^N =\infty.$$ Which will prove the non-uniformly convergence for your series.