Asymptotic behavior of a sequence based on a subsequence II

Let $\{a_{n}\}$ be a non-increasing sequence of positive numbers. if for some positive integers $l,p$ and $R>1$ we have $a_{(ln)^{p}}=O(R^{-n})$ as $n\to\infty$, what can we say about the behavior of $a_{n}$? tkx!!

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Without further information, such as monotonicity, we cannot say anything further. –  sos440 Aug 22 '12 at 13:45

If you consider $a_{i+1}\le a_i$ then

$$f(n)=-\frac{1}{l}\sqrt[p]{n}$$

$$a_n=O(R^{f(n)})$$

DETAILS :

consider $m$ such that $(lm)^p\le n \lt l(m+1)^p$, then $m=\lfloor\,f(n)\rfloor$

As $a_{(ln)^p}=O(R^{-n})$, there exists $k>0$ such that $a_{(ln)^p}\le k(R^{-n})$

So for any $n$ (and $R\ge 1$), $$a_n<a_{l(m+1)^p}\le k.R^{-(m+1)}\le k.R^{-f(n)}$$

and if $(R<1)$

$$a_n<a_{l(m+1)^p}\le k.R^{-(m+1)}\le k.R.R^{-f(n)}$$

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Could you give some details ? Please? –  Dubglass Aug 23 '12 at 3:04