# Prove that two Cauchy sequences are convergent.

Prove that $$x_n+y_n$$ is Cauchy without using the theorem: A sequence is Cauchy iff it converges to some point.

My attempt:

Assume that $$x_n$$ and $$y_n$$ are convergent. That is by definition $$\vert x_n - a \vert \lt \frac{\epsilon}{2}$$ $$\vert y_n - a \vert \lt \frac{\epsilon}{2}$$

$$\vert x_n -a \vert + \vert y_n - a \vert \lt \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$ By definition of convergence $$x_n + y_n$$ is Cauchy

• Did you mean to write $y_n-a$ in the third stage? – Mohammad Zuhair Khan Sep 30 at 17:03
• yes my mistake. – K. Gibson Sep 30 at 17:03
• We don't need to assume equal limit for $x_n$ and $y_n$. – user Sep 30 at 17:06
• what do you mean? – K. Gibson Sep 30 at 17:10
• @K.Gibson Can you please clarify: are you proving that the sum of two Cauchy sequences is Cauchy, or that the sum of two convergent sequences is convergent? As you say you are not allowed to use the equivalence of those two conditions, for the purpose of the proof we must first know which of those two conditions is at play here. – Stinking Bishop Sep 30 at 17:11

We have that

• $$\vert x_n - L_1 \vert \lt \frac{\epsilon}{2}$$

• $$\vert y_n - L_2 \vert \lt \frac{\epsilon}{2}$$

then by triangle inequality

$$\vert x_n+y_n-(L_1+L_2)\vert\le \vert x_n -L_1 \vert + \vert y_n - L_2 \vert \lt \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$

and $$x_n+y_n$$ converges.

Let's assume that $$(x_n)$$ and $$(y_n)$$ are two Cauchy sequences. We will prove that $$(x_n+y_n)$$ is Cauchy.

Pick $$\varepsilon>0$$. For this $$\varepsilon$$, there is $$n_1\in\mathbb N$$ such that:

$$|x_m-x_n|<\frac{\varepsilon}{2}\text{ for }m,n\ge n_1$$

and also there is $$n_2\in\mathbb N$$ such that:

$$|y_m-y_n|<\frac{\varepsilon}{2}\text{ for }m,n\ge n_2$$

Now let $$n_0$$ be the maximum of $$n_1, n_2$$. Then, for $$m,n\ge n_0$$ we obviously have $$m,n\ge n_1, n_2$$, so the following is valid (using the triangle inequality):

$$\begin{array}{rcl}|(x_m+y_m)-(x_n+y_n)|&=&|(x_m-x_n)+(y_m-y_n)|\\&\le&|x_m-x_n|+|y_m-y_n|\\&\lt&\frac{\varepsilon}{2}+\frac{\varepsilon}{2}\\&=&\varepsilon\end{array}$$

Take $$\varepsilon>0$$. There are $$N_1,N_2\in\mathbb N$$ such that$$n\geqslant N_1\implies\lvert x_n-a\rvert<\frac\varepsilon4\text{ and }n\geqslant N_2\implies\lvert y_n-b\rvert<\frac\varepsilon4.$$Let $$N=\max\{N_1,N_2\}$$. Then, if $$m,n\geqslant N$$, you have\begin{align}\bigl\lvert(x_m+y_m)-(x_n+y_n)\bigr\rvert&=\lvert x_m-a+a-x_n+y_m-b+b-y_n\rvert\\&\leqslant\lvert x_m-a\rvert+\lvert x_n-a\rvert+\lvert y_m-b\rvert+\lvert y_n-b\rvert\\&<\frac\varepsilon4+\frac\varepsilon4+\frac\varepsilon4+\frac\varepsilon4\\&=\varepsilon.\end{align}