# Determine rate of convergence for $a_n=b(1-\sqrt{3})^n$, where $b \in \mathbb{R}$

I have to determine the order of the rate of convergence for $$a_n=b(1-\sqrt{3})^n$$, where $$b \in \mathbb{R}$$. The sequence is obviously converging to $$0$$, and I know the order is at least linear, since we have:

$$\frac{\left|a_{n+1}-0\right|}{\left|a_{n}-0\right|} = \left|\frac{b(1-\sqrt{3})^{n+1}}{b(1-\sqrt{3})^{n}}\right| = \left|\frac{(1-\sqrt{3})^{n+1}}{(1-\sqrt{3})^{n}}\right| = |1-\sqrt{3}| < 1$$

and so there is a constant $$c<1$$ and an integer N such that:

$$|a_{n+1}-0| \leq (\sqrt{3}-1) |a_n-0| \quad (n\geq N)$$

and so the rate of convergence is at least linear. But is it superlinear?

I am not sure how to determine this, but here's my try: If it's the case, then there exists a sequence $$\epsilon_n$$ tending to $$0$$ and an integer N such that:

$$|a_{n+1}-0| \leq \epsilon_n |a_n-0| \quad (n\geq N)$$

So I think this is false because we know that:

$$\frac{\left|a_{n+1}-0\right|}{\left|a_{n}-0\right|} = |1-\sqrt{3}| \rightarrow |1-\sqrt{3}| \quad \text{for} \quad n \rightarrow \infty$$

which certainly does not converge to $$0$$.

Is this the way to deal with problems like this, and are my thoughts/solution above correct?

If $$\epsilon_n:=a_n-L$$ with $$L:=\lim_{n\to\infty}a_n$$ and constants $$K,\,p>0$$ exist with $$\lim_{n\to\infty}\frac{|\epsilon_{n+1}|}{|\epsilon_n|^p}=K$$, the convergence is superlinear if $$p>1$$ (exercise: prove $$p\ge 1$$), linear if $$p=1$$ and $$K<1$$, and sublinear if $$p=K=1$$. In your case $$p=1,\,K=\sqrt{3}-1$$, so the convergence is linear.
• Okay. So for example, in the case $p=2$, we have $\frac{\left|a_{n+1}-0\right|}{\left|a_{n}-0\right|^2} = \frac{1}{\left|b(1-\sqrt{3})^{n-1}\right|} \rightarrow \infty \quad \text{for} \quad n \rightarrow \infty$, and so it can't be superlinear? Oct 7, 2018 at 20:53
• @Jazzman Yes; at most one $p$ exists satisfying $0<K<\infty$, so since $p=1$ works $p=2$ doesn't.