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I know that the rate of convergence of a sequence is unique. But I do not know why. Please prove that the rate of convergence of a sequence is unique.

Rate of Convergence

Rate of Convergence Definition

Behrooz Fard
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Suppose $e_n > 0$ and $e_n \rightarrow 0$ for $n \rightarrow \infty$ and $n \in \mathbb{N}$. Let us assume that the order of convergence is both $p_1\geq1$ and $p_2 \geq 1$. We have to show that $p_1 = p_2$. By assumption we have $$ \frac{e_{n+1}}{e_n^{p_i}} \rightarrow c_i > 0, \quad n \rightarrow \infty, \quad n \in \mathbb{N}.$$ Without loss of generality, we have $p_1 < p_2$. If $p_1 = 1$, then the demand is $c_1 \in (0,1)$, but this will not be important here.

It follows that $$ \frac{e_{n+1}}{e_n^{p_1}} = \frac{e_{n+1}}{e_n^{p_2}} e_n^{p_2 - p_1} \rightarrow c_2 \cdot 0 = 0, \quad n \rightarrow \infty, \quad n \in \mathbb{N}$$ because $p_2 - p_1>0$ and $e_n$ tends to zero. This shows that the order is not $p_1$. We have a contradiction. Hence $p_1 = p_2$.

Carl Christian
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    @bezi The order is ultimately more significant than than rate, unless the order is $1$, in which case a small rate ensures faster convergence. In my native language (Danish), "rate" is valid word with connotations that very close to speed which in our setting is closely tied to order. It requires an effort for me to distinguish between the terms "rate" and "order". Since the rate is literally a limit, it did not even occur to me that you meant anything but "order". – Carl Christian Oct 19 '20 at 19:44
  • I have actually find in some cases that rates can have a rather significant contribution as well as the order. This is particularly the case when dealing with bracketing methods, where one endpoint stalls. In this situation, the rate is usually proportional to how close this stalled point is to the root, which for single or double precision may have a significant impact. – Simply Beautiful Art Oct 19 '20 at 21:39