Proof that superlinearly convergent sequence converges faster than linearly convergent sequence

Question

Given real sequences $(a_n)_{n\in\mathbb{N}}$ and $(b_n)_{n\in\mathbb{N}}$, both converging to the same limit $A$ and such that $|a_n-A|\neq 0$ and $|b_n-A|\neq 0$ for every $n$ sufficiently large, we say that $(a_n)_{n\in\mathbb{N}}$ converges faster to $A$ than $(b_n)_{n\in\mathbb{N}}$ if $$\lim_{n\to\infty}\frac{|a_n-A|}{|b_n-A|}=0.$$

If a real sequence $(c_n)_{n\in\mathbb{N}}$ converging to a limit $C$ satisfies the condition that $$\lim_{n\to\infty}\frac{|c_{n+1}-C|}{|c_n-C|}=\lambda$$ for some $\lambda\in(0,1)$, we say that $(c_n)_{n\in\mathbb{N}}$ converges linearly to $C$, and, if a sequence $(c^\prime_n)_{n\in\mathbb{N}}$ converges to a limit $C^\prime$ and satisfies the condition $$\lim_{n\to\infty}\frac{|c^\prime_{n+1}-C^\prime|}{|c^\prime_n-C^\prime|}=0,$$ we say that $(c^\prime_n)_{n\in\mathbb{N}}$ converges superlinearly to $C^\prime$.

Question: Given the definitions above, is it generally true that a sequence converging superlinearly to a given limit converges faster than a squence converging linearly to the same limit? How do I prove that?

Answer 1

Wlog the limit is $0$. Pick $N$ so that the limit in the defintion of linear convergence holds to within $\frac{\lambda}{2}$ past $N$, it then follows that

$$|a_{n+k}| \geq h^k |a_n|$$

for all $n \geq N$, with $h := \frac{\lambda}{2}$.

Meanwhile, with $n_{\delta} \geq N$ picked large enough also so that the limit in the defintion of superlinear convergence held to within $\delta$ past $n_\delta$, we obtain

$$|b_{n_{\delta} + k}| \leq \delta^k |b_{n_{\delta}}|.$$

So

$$\frac{|b_{n_\delta + k}|}{|a_{n_\delta + k}|} \leq \left(\frac{\delta}{h}\right)^k \frac{|b_{n_\delta}|}{|a_{n_\delta}|},$$

which shows the desired conclusion once $\delta$ is chosen less than $h$.