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
-1
$\begingroup$
sequences-and-series
asymptotics
-
0Without further information, such as monotonicity, we cannot say anything further. – 2012-08-22
1 Answers
1
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
-
0Could you give some details ? Please? – 2012-08-23