Let $(x_n)$ and $(y_n)$ be Cauchy sequences. Give a direct proof that $(2x_n + y_n)$ is a Cauchy sequence.

2017-12-03 03:04:25

It is my first time being stuck on a problem in my basic analysis book. The book I am using basically just talks about Cauchy's life and gives a couple definitions and no propositions. I am confused on where to start and was wondering if someone could give me some pointers on how to prove this.

Let $\epsilon>0$. The fact that $(x_n)_{n\in\mathbb{N}}$ is a cauchy sequence gives you some natural number $N_1\in \mathbb{N}$ such that forall $n,m>N_1$ it holds $|x_n-x_m|<\epsilon$ for the distance. Because $(y_n)_n$ is a cauchy sequence too you get some $N_2$ with a similar property for $(y_n)_n$. Now let $N:=max(N_1,N_2)$. Then for all $n,m>N$ it holds that $|(2x_n+y_n)-(2x_m+y_m)|<\ldots$. Hint: Use the triangular inequality.

  • Let $\epsilon>0$. The fact that $(x_n)_{n\in\mathbb{N}}$ is a cauchy sequence gives you some natural number $N_1\in \mathbb{N}$ such that forall $n,m>N_1$ it holds $|x_n-x_m|<\epsilon$ for the distance. Because $(y_n)_n$ is a cauchy sequence too you get some $N_2$ with a similar property for $(y_n)_n$. Now let $N:=max(N_1,N_2)$. Then for all $n,m>N$ it holds that $|(2x_n+y_n)-(2x_m+y_m)|<\ldots$. Hint: Use the triangular inequality.

    2017-12-03 05:41:15