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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!?

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    $\begingroup$ 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). $\endgroup$ – Nicolas Feb 8 at 20:37
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    $\begingroup$ 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 $M \in \mathbb{N}$. $\endgroup$ – stressed out Feb 8 at 21:29
  • $\begingroup$ The idea is that, starting at some index, any two numbers in that sequence are "close enough" to each other, how close? epsilon-close! $\endgroup$ – Wesley Strik Feb 9 at 10:06
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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.

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