Tail of Convergent Sequence
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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 $a_n \to a$.
$\blacksquare$