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?


1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$ Oct 7, 2018 at 20:53
  • $\begingroup$ @Jazzman Yes; at most one $p$ exists satisfying $0<K<\infty$, so since $p=1$ works $p=2$ doesn't. $\endgroup$
    – J.G.
    Oct 7, 2018 at 21:16

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