Given two convergent sequences $\{a_n\}$ and $\{b_n\}$ with the same limit point $x^*$, I want to show the convergence rate of the sequence of the diameter $\{b_n-a_n\}$. Suppose $a_n\leq x^*$ for all $n=1,2,3,\ldots$ and $b_n\geq x^*$ for all $n=1,2,3,\ldots$. Moreover, $\{a_n\}$ converges quadratically and $\{b_n\}$ superlinearly converges to the limit point. In other words, I have $\lim_{n\rightarrow\infty}\frac{a_{n+1}-x^*}{(a_n-x^*)^2}=\mu$, where $\mu>0$ and $\lim_{n\rightarrow\infty}\frac{b_{n+1}-x^*}{b_n-x^*}=0$.

I guess the sequence of the diameter $\{b_n-a_n\}$ will converge superlinearly. However, I don't know how to prove algebraically. Any help and comment will be greatly appreciated. Thanks!


1 Answer 1


In particular, $(a_n)$ is superlinearly convergent to $x^*$, and $a_n\leq x^*\leq b_n$, so \begin{align*} \dfrac{b_{n+1}-a_{n+1}}{b_n-a_n}&=\dfrac{(b_{n+1}-x^*)+(x^*-a_{n+1})}{(b_n-x^*)+(x^*-a_n)}\\ &=\dfrac{b_{n+1}-x^*}{(b_n-x^*)+(x^*-a_n)}+\dfrac{x^*-a_{n+1}}{(b_n-x^*)+(x^*-a_n)}\\ &\leq\dfrac{b_{n+1}-x^*}{b_n-x^*}+\dfrac{x^*-a_{n+1}}{x^*-a_n}\to 0\qquad\text{as}\qquad n\to\infty \end{align*} (the inequality was obtained from the fact that $\dfrac{1}{x+y}\leq\dfrac{1}{x}$ and $\dfrac{1}{y}$ whenever $x,y\geq 0$: just use $x=(b_n-x^*)$ and $y=(x^*-a_n)$).

Also, the left hand-side above is positive, hence $\left|\dfrac{b_{n+1}-a_{n+1}}{b_n-a_n}\right|\to 0$ as $n\to \infty$, and this means that $(b_n-a_n)$ is superlinearly convergent.

  • $\begingroup$ how did you get the inequality? $\endgroup$
    – chp61
    Sep 4, 2014 at 14:02
  • $\begingroup$ @chp61 I added some details to the answer $\endgroup$ Sep 4, 2014 at 14:30
  • $\begingroup$ Thanks a lot. It makes sense :) $\endgroup$
    – chp61
    Sep 5, 2014 at 12:56

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