Absolute and conditional convergence of function series

Question

I have a problem. I have to explore absolute and conditional convergence of this function series

Unfortunately. I didn't find in my reference any words about "absolute and conditional", Instead I've just seen "uniform convergence, amount and balance" of function series. I've jusy started to learn it. Help me find out the steps of solving and solve my example.

Thank You so much!

Answer 1

This series is absolutely convergent:

$$\big| \sum_{n=1}^{ \infty}3^n \sin \left( \frac{x}{5^n} \right) \big| \le \sum_{n=1}^{ \infty}3^n \sin \left( \frac{|x|}{5^n} \right) \le \sum_{n=1}^{ \infty}3^n \frac{|x|}{5^n}=\\|x|\sum_{n=1}^{ \infty} \left( \frac{3}{5} \right)^n=|x| \left(\frac{1}{1-3/5}-1 \right)= \frac{3|x|}{2}$$

by summing the geometric series and noting that $-|x| \le \sin(x) \le |x|\ \forall x \in \mathbb{R}$.

Answer 2

Given $\sum_{n \ge 1} a_n$, if $\sum_{n \ge 1} |a_n|$ converges we call the series absolutely convergent. If $\sum_{n \ge 1} a_n$ converges but $\sum_{n \ge 1} |a_n|$ diverges we call the series conditionally convergent.

Hints about your problem:

1. Show that $\lim_{n \to \infty}\frac{3^n| \sin(\frac{1}{4^n})|}{3^n | \sin(\frac{x}{5^n})|}= \infty$ . (I chose $4$ here just because it is between $3$ and $5$. Instead you can in fact choose any number between 3 and 5 without 3 and 5, it's not a big deal)

2. Using 1 deduce that for every given $x$ there exists $n_0$ such that for all $n>n_0$ we have $3^n | \sin(\frac{x}{5^n})| \le 3^n | \sin(\frac{1}{4^n})|$.

3. Show that $3^n| \sin(\frac{1}{4^n})| \sim_{\infty} \frac{3^n}{4^n}$ .

4. Give the conclusion.