# Show directly that if $\{s_n\}$ is a Cauchy sequence then so is $\{|s_n|\}$. Conclude that $\{|s_n|\}$ converges whenever $\{s_n\}$ converges.

Show directly that if $$\{s_n\}$$ is a Cauchy sequence then so is $$\{|s_n|\}$$. From this conclude that $$\{|s_n|\}$$ converges whenever $$\{s_n\}$$ converges.

Let $$\{s_n\}$$ be a Cauchy sequence. Then by definition, for any given $$\varepsilon>0$$ there exists $$m>0$$ such that $$|s_n-s_m|<\varepsilon$$ for all $$n\geq m$$. Then we have $$||s_n|-|s_m||\leq|s_n-s_m|$$ Therefore, from the definition $$||s_n|-|s_m||\leq|s_n-s_m|<\varepsilon$$ for all $$n\geq m$$. Hence, $$\{|s_n|\}$$ is a Cauchy sequence.

And then to prove that convergence of $$\{s_n\}$$ implies the convergence of $$\{|s_n|\}$$:

Let $$\varepsilon>0$$. If $$\{s_n\}$$ converges to $$L$$, then there exists $$N$$ such that $$|s_n-L|<\varepsilon$$, whenever $$n\geq N$$. Hence, for $$n\geq N$$, we have $$||s_n|-L|\leq |s_n-L|<\varepsilon$$. Thus $$\{|s_n|\}$$ converges to $$|L|$$.

That's how I proved but I'm not sure if I possibly made some mistakes or missed some steps!?

• Hint: Reverse triangle inequality (en.wikipedia.org/wiki/…). From there, it suffices to show that every Cauchy sequence converges (if you haven't shown that already). – Nicolas Feb 8 at 20:37
• Your proof is close to being correct but your definition of a sequence being Cauchy is not. Instead of $n\geq m$, it should be $\forall n,m \geq M$ for some $M \in \mathbb{N}$. – stressed out Feb 8 at 21:29
• The idea is that, starting at some index, any two numbers in that sequence are "close enough" to each other, how close? epsilon-close! – Wesley Strik Feb 9 at 10:06

If we are dealing with a complete metric space, we know that every Cauchy sequence is convergent to a limit in that metric space. So if you prove that the sequence is $$\{|s_n| \}$$ is Cauchy, indeed by using the reverse triangle inequality, we automatically get that it is a convergent sequence.