# Tail of Convergent Sequence

## Theorem

Let $\left\langle{a_n}\right\rangle$ be a real sequence.

Let $m \in \N$ be a natural number.

Let $a \in \R$ be a real number.

Then:

$a_n \to a$
$a_{n + m} \to a$

## Proof

### Necessary Condition

Suppose that $a_n \to a$.

Then for each $\epsilon > 0$, there exists $N \in \N$ such that:

$\size {a_n - a} < \epsilon$ for $n \ge N$.

Set:

$N^\ast = \max \set {1, N - m}$

Then for $n \ge N^\ast$, we have $n \ge N - m$ and so $n + m \ge N$.

Then:

$\size {a_{n + m} - a} < \epsilon$ for $n \ge N^\ast$.

So $a_{n + m} \to a$.

$\Box$

### Sufficient Condition

Suppose that $a_{n + m} \to a$ for all $m \in \N$.

Then in particular $a_{n + 1} \to a$.

So for each $\epsilon > 0$ there exists $N \in \N$ such that:

$\size {a_{n + 1} - a} < \epsilon$ for $n \ge N$.

Then:

$\size {a_n - a} < \epsilon$ for $n \ge N + 1$.

So $a_n \to a$.

$\blacksquare$