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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!

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  • $\begingroup$ 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
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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.

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