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I came accross two different definitions of order convergence, one (Def 1) given by my professor (I can't find any same ones elsewhere), and one (Def 2) which can be found in any materials available on the Internet, such as here.

$\{p_k\}$ is an $\alpha- $order convergent sequence, where $p_k \rightarrow p$, iff

Def 1 \begin{equation} \exists \lambda>0,\ \exists k^* \ge 0,\ \forall k\ge k^*,\ ||p_{k+1}-p|| \le \lambda ||p_k-p||^{\alpha} \end{equation}

Def 2 \begin{equation} \exists \lambda>0,\ \exists \alpha > 0, \lim_{k\to+\infty} \frac{||p_{k+1}-p||}{||p_k-p||^{\alpha}} = \lambda \end{equation}

How are the two equivalent to each other?

user36706
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1 Answers1

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The second definition isn’t quite right: you’re defining $\alpha$-order convergence, so $\alpha$ should be a free variable. It should say simply that there is a $\lambda>0$ such that $\lim\limits_{k\to\infty}\frac{\|p_{k+1}-p\|}{\|p_k-p\|^\alpha}=\lambda$.

And it is definitely stronger than the first definition. Let

$$p_k=\left(\frac12\right)^{2^k}$$

for $k\in\Bbb N$; clearly $\langle p_k:k\in\Bbb N\rangle$ converges to $0$, and

$$\left|\left(\frac12\right)^{2^{k+1}}\right|\le\left|\left(\frac12\right)^{2^k}\right|$$

for all $k\in\Bbb N$, so by the first definition the sequence is $1$-convergent with $\lambda=1$. It is also $2$-convergent:

$$\left|\left(\frac12\right)^{2^{k+1}}\right|=\left|\left(\frac12\right)^{2^k}\right|^2\,.$$

However,

$$\frac{\left|\left(\frac12\right)^{2^{k+1}}\right|}{\left|\left(\frac12\right)^{2^k}\right|}=\left(\frac12\right)^{2^{k+1}-2^k}=\left(\frac12\right)^{2^k}$$

converges to $0$ as $k$ increases without bound, so the sequence is not $1$-convergent by the second definition. It is $2$-convergent by the second definition:

$$\frac{\left|\left(\frac12\right)^{2^{k+1}}\right|}{\left|\left(\frac12\right)^{2^k}\right|^2}=1$$

for all $k\in\Bbb N$.

It’s not hard to check that if a sequence is $\alpha$-convergent by the second definition, it is $\beta$-convergent by the first definition whenever $0<\beta\le\alpha$.

Brian M. Scott
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