# Cauchy sequence Convergence on real numbers

Suppose $\{a_n\}$ and $\{b_n\}$ are Cauchy sequences of $\mathbb{R}$. Show that $\{a_n+b_n\}$ is also Cauchy sequences.

My proof is like this. Since this is a Cauchy sequences of $\mathbb{R}$, $\{a_n\}$ and $\{b_n\}$ converges. Thus, $\{a_n+b_n\}$ converges. Since every convergent sequences is Cauchy sequences, $\{a_n+b_n\}$ is Cauchy sequence.

Is it okay to prove like this instead of using definition of Cauchy sequence to prove it?

• You haven't really proved anything without the proper definitional manipulation... You've simply said that the sequence just is convergent... – Isky Mathews Apr 24 '18 at 19:15
• How do you know that if $\{a_i\}$ converges and $\{b_i\}$ converge, then it is true that $\{a_i + b_i\}$ converge? If you have justification for that then you can state it if you site the justification. If you don't have justification then you must prove it. As it stands, no, that is not a proof. That is merely restating what needs to be proven into an equivalent statement that will also need to be proven. – fleablood Apr 24 '18 at 19:30

I will say, though, that you've used a really heavy piece of machinery (that any Cauchy sequence in $\mathbb{R}$ converges) when it is completely unnecessary for the problem at hand.
It is very straight-forward to prove this fact directly. Take an $\epsilon>0$, and use the fact that $(a_n)$ and $(b_n)$ are Cauchy to show that both $\lvert a_n-a_m\rvert$ and $\lvert b_n-b_m\rvert$ can be made small for all $n,m\geq N$, for some $N\in\mathbb{N}$. Then use the triangle inequality to consider $\lvert (a_n+b_n)-(a_m+b_m)\rvert$ when $n,m$ exceed that same $N$.