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Here's the problem: $k>1,k\in\mathbb{N}.$ Given $\ a_0>0,\forall n\in \mathbb{N},a_{n+1}=a_n+a_{n}^{-\frac{1}{k}}.$ Figure out $\lim\limits_{n\to \infty}\frac{(a_n)^{k+1}}{n^k}.$

I have no idea of dealing with the relation.It makes sense to say $a_n$ grows rather slowly,every time by $-\frac{1}{k}$times of itself.Thus $k+1$times makes $a_n$ grows like what $n^k$does.But how to make it to the ground?Are there any techniques to deal with such relations?Anything about it would be highly appreciated.

shdvt
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  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Oct 07 '21 at 13:40
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    I don't have my copy here, but Asymptotic Methods in Analysis by de Bruijn might have something useful. – Matthew Towers Oct 07 '21 at 13:42
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    Attempt a numerical simulation for example with $k=2$ to obtain a certain number of elements $a_n$ and using thew to have an idea of the limit of $a_n^{k+1}/n^k$, because, if you don't know what is to be proven... you will not progress... – Jean Marie Oct 07 '21 at 21:14
  • A very similar recurrent sequence is considered here – Jean Marie Oct 07 '21 at 21:21

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Thanks for all your comments above.With the assumption that $a_n\sim \alpha n^{\frac{k}{k+1}},$back to $a_{n+1}^{k+1}=\sum\limits_{r=0}^{k+1}\binom{k+1}{r}(\alpha n^{\frac{k}{k+1}})^{\frac{(k+1)(k-r)}{k}}$we extract the factor of $n^{k}$and get $\alpha$:$k \alpha^{k+1}=(k+1)\alpha^{\frac{(k+1)(k-1)}{k}},\alpha=(\frac{k+1}{k})^\frac{k}{k+1}.$Thus we have $a_n^{k+1}\sim (1+\frac{1}{k})^k n^k$,which is a quite beautiful result.

shdvt
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  • And I've varified the answer by programming...The limits when $k=2,...,10$ are 2.25 2.37037 2.44141 2.48832 2.52163 2.5465 2.56578 2.58117 2.59374,which are just the second to the tenth number of $(1+\frac{1}{k})^k$. – shdvt Oct 08 '21 at 05:00