Suppose $(a_n)$ is a Cauchy sequence of rationals and $b_n = a_n + a_{n+1}$.

Question

Suppose $(a_n)$ is a Cauchy sequence of rationals and $b_n = a_n + a_{n+1}$. Prove that $(b_n)$ is also a Cauchy sequence. You will want to use a different $N$ for the same $ε$ in each case. This is from "Tools of the Trade" by Paul Sally Jr., exercise $3.5.13$.

Answer 1

Let $\varepsilon>0$, since $\{a_n\}_{n=1}^{\infty}$ is a Cauchy sequence, there exist $N_{\varepsilon}$ such that $$m,n\ge N_{\varepsilon}\qquad\implies\qquad|a_m-a_n|<\varepsilon/2$$ Then \begin{align} m,n\ge N_{\varepsilon}\qquad\implies\qquad|b_m-b_n|&=|a_{m+1}+a_{m}-a_{n+1}-a_{n+1}|\\ &\le|a_{m+1}-a_{n+1}|+|a_m-a_n|\\ &<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}\\ &=\varepsilon \end{align}

Answer 2

Hint:

\begin{align*} |b_n-b_m| & =|(a_n+a_{n+1})-(a_m+a_{m+1})|\\ & =|(a_n-a_{m})+(a_{n+1}-a_{m+1})|\\ & \leq |a_n-a_{m}|+|a_{n+1}-a_{m+1}| &&(\text{by triangle inequality}) \end{align*} You are given that $\{a_n\}$ is Cauchy: so For every $\epsilon >0$, there exists a $K$ such that for all $m,n >K$, $$|a_n-a_{m}| \leq \epsilon.$$ Now use this to get bounds on $|b_n-b_m|$.