# Limits, n, n+1

1. ### Tclack

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.

2. ### CompuChip

4,297
This is about as informal as they get, but ... if you delete from the second limit your first element, you shift everything down by one and you get the same limit again.

3. ### Tclack

37
I thought of something.
As X_n goes to infinity, it passes X_(n+1)

4. ### HallsofIvy

40,780
Staff Emeritus
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.)

5. ### Dickfore

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?