Proof of Constant Sequences and Index-Shifted Sequences of Real Numbers

1. (Constant sequences.) Let $\varepsilon>0$ and take $N=1$. For every $n\in\mathbb{N}$ with $n\ge N$ the $n$th term of the constant sequence is $c$, and

$$|c-c|=|0|=0<\varepsilon$$

by Properties of the Absolute Value in an Ordered Field. Hence the constant sequence converges to $c$.

2. (Index shift.) Suppose $(a_n)$ converges to $L$, and let $\varepsilon>0$. Choose $N\in\mathbb{N}$ such that $|a_m-L|<\varepsilon$ for every $m\in\mathbb{N}$ with $m\ge N$. Let $n\in\mathbb{N}$ with $n\ge N$. By clause 6 of Properties of the Order on the Natural Numbers, $n<n+1$ in $\mathbb{N}$; by clause 1 of that lemma this gives $n\le n+1$, and combining with $N\le n$ through the transitivity in the same clause gives $N\le n+1$. Therefore

$$|b_n-L|=|a_{n+1}-L|<\varepsilon .$$

Hence $(b_n)$ converges to $L$.