Prove that two Cauchy sequences are convergent.

Question

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

Answer 1

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.

Answer 2

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}

Answer 3

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):

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