# Limits, n, n+1

by Tclack
Tags: infinity, limit
 P: 37 So I came across the statement: Since $x_n -> \inf$ then $x_n_+_1 -> \inf$ This is very basic, But I'm already into recursive formulas for infinite series, so I should know why this is true. Does anyone have a small proof. An informal one will do.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,556 Limits, n, n+1 I'm not sure what that means! For any n, n< n+1. But I think you mean that, for fixed N, eventually, n> N+1. A little more precisely, if $\{a_n\}$ converges to A, then, for any $\epsilon> 0$, there exist N such that if n> N, $|a_n- A|< \epsilon$. Now, if $b_n= a_{n+1}$, for any $\epsilon> 0$, take N'= N-1 where N is the number, above, for that same $\epsilon$. Then if n> N' , n+1> N'+1= N so $|b_n- A|= |a_{n+1}- A|< \epsilon$, showing that $\{b_n\}$ also converges to A. (Roughly speaking, $\{a_n\}$ and $\{a_{n+1}\}$ are really the same sequence, just with the "numbering" altered slighly. Of course, they have the same limit.)
 P: 3,014 You need to know what subsequences are: You choose a strictly monotonically increasing sequences of natural numbers: $$n_{k}, n_{k} \in \mathbb{N}, n_{k + 1} > n_{k}, \; k = 0, 1, \ldots$$ Then, a subsequence of the sequence $\{x_{n}\}$ is defined as: $$\tilde{x}_{k} \equiv x_{n_{k}}$$ The (informal) theorem you will need to remember is: Any subsequence has the same convergence properties and the same limit if convergent as its sequence. How would you choose your subsequence?