Order of Convergence Implies Convergence

Theorem

Let $\sequence {x_n}$ be an real sequence that converges to $\alpha$ with order $p$, where $p \ge 1$.

Then, $\sequence {x_n}$ converges to $\alpha$.

Proof

By definition of order of convergence, there exist a sequence $\sequence {\epsilon_n}$ and $c > 0$ such that:

$\size {x_n - \alpha} \le \epsilon_n$

$\ds \lim_{n \to \infty} \frac {\epsilon_{n + 1}} {\epsilon_n^p} = c$

$p = 1 \implies c < 1$

Case $p = 1$

Suppose $p = 1$.

Then, we have:

$\ds \lim_{n \to \infty} \frac {\epsilon_{n + 1}} {\epsilon_n} = c < 1$

By definition of limit, there exists some $N \in \N$ such that:

$\forall n > N: \size {\dfrac {\epsilon_{n + 1}} {\epsilon_n} - c} < \dfrac {1 - c} 2$

In particular, for $n > N$:

$\dfrac {\epsilon_{n + 1}} {\epsilon_n} < c + \dfrac {1 - c} 2 = \dfrac {c + 1} 2 < 1$

Let $c' = \dfrac {c + 1} 2 < 1$.

Then:

$\epsilon_{n + 1} < c' \epsilon_n$

It follows that, for every $n \in \N$:

$\epsilon_{N + 1 + n} \le \paren {c'}^n \epsilon_{N + 1}$

By Sequence of Powers of Number less than One, Multiple Rule for Real Sequences, Squeeze Theorem for Real Sequences:

$\sequence {\epsilon_n}$ is a null sequence

Therefore, by Corollary to Squeeze Theorem for Real Sequences:

$x_n \to \alpha$

$\Box$

Case $p > 1$

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