The question is,

If $x_k \to L$, does it always exist some number $p$ such that the limit $\mathop {\lim }\limits_{k \to \infty } \frac{{|{x_{k + 1}} - L|}}{{|{x_k} - L{|^p}}}$ exist?

I think this is true, but I have trouble proving it. Hope someone can help. Thank you!

  • $\begingroup$ if the sequence is monotonically converging to $L$, then $p=0$, or any $p \in [0;1]$ $\endgroup$ – reuns Jan 13 '16 at 2:55
  • $\begingroup$ You need at least $x_n \ne L$ or else the conjecture fail for the constant sequence L, L, L, ... $\endgroup$ – BigbearZzz Jan 13 '16 at 7:38

This is not true in general if $p > 0$. Without the lost of generality we can assume that $L =0$. For non triviality, let's assume that $x_n \ne 0$ for all $n \in \Bbb N$. Now, consider the sequence $$ x_n= \begin{cases} \frac1n, &\text{$n$ is odd} \\ \frac1{2^n}, &\text{$n$ is even} \end{cases} $$ You can verify that $\frac{|x_{k+1}|}{|x_k|^p}$ has two subsequences with the asymptotic behavior of $\frac{n^p}{2^n}$ and $\frac{2^{np}}{n}$, respectively. Clearly the first one converges to $0$ but the later diverges to infinity as $n\to \infty$, as long as $p>0$.

However, if we allow $p\le 0$ then the limit in question always exists if we let $p=-1$ so that $\mathop {\lim }\limits_{k \to \infty } \frac{{|{x_{k + 1}} - L|}}{{|{x_k} - L{|^p}}}=\lim_{k \to \infty}|x_{k+1}-L||x_k-L|=0$.

The case where $p=0$ is also trivial.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.