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!!
Asymptotic behavior of a sequence based on a subsequence II
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sequences-and-series
asymptotics
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0Without further information, such as monotonicity, we cannot say anything further. – 2012-08-22
1 Answers
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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 $R\ge 1$), $$a_n and if $(R<1)$ $$a_n
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0Could you give some details ? Please? – 2012-08-23