- #1

Eclair_de_XII

- 1,083

- 91

## Homework Statement

"Prove that if ##\{a_n\}## converges to ##A##, then ##\{|a_n|\}## converges to ##|A|##. Is the converse true?"

## Homework Equations

Triangle inequality: ##|x+y|≤|x|+|y|##

## The Attempt at a Solution

I think I solved the first part...

"If ##\{a_n\}## converges to ##A##, then there must exist an ##N∈ℕ## such that for each ##\epsilon>0##, ##|a_n-A|<\epsilon## for all ##n≥N##. In turn, ##|a_n|=|a_n-A+A|≤|a_n-A|+|A|##, which implies that ##|a_n-A|≥|a_n|-|A|##. Taking the absolute values of both terms gives: ##|a_n-A|≥| |a_n|-|A| |##. Thus, ##\epsilon>|a_n-A|≥| |a_n|-|A| |## for all ##\epsilon>0## and for all ##n≥N##."

For the second part, I basically came up with a counterexample:

"Define ##\{a_n\}## by ##\{a_n:a_n=-1,∀n∈ℕ\}##. We note that ##\{a_n\}## does not converge to ##1##, but ##\{|a_n|:|a_n|=|-1|,∀n∈ℕ\}## does converge to ##1##."

But I don't think this will cut it... What I want to find now, is an ##\epsilon>0## such that ##\epsilon>| |a_n|-|A| |## but ##|a_n-A|>\epsilon##. Hence, I will need to find a number ##\epsilon## such that ##|a_n-A|>\epsilon>| |a_n|-|A| |##, to show that the converse is not true. I will need to find an appropriate choice of ##\epsilon## that satisfies these conditions, in other words.