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I was thinking about the following problem:

Consider a sequence $a_n \to 0, a_n >0$ which is monotone, i. e. $a_n\ge a_{n+1}$.

Now suppose for every $C>0$ there is a subsequence $n(k)$ such that $a_{n(k)}>C\ f(n(k))$, where f is a function fulfilling $f(x)\to 0 \ (x\to\infty)(e.g. f(x)=x^{-1})$.

Can you conclude now $a_{n}>C'\ f(n)$ for all $n$?

I think so, because each subsequence $a_{n(1)-i},a_{n(2)-i},\ldots$ can not decay faster than $f$. Even when the difference $|n(k)-n(k-1)|$ is not bounded, you can choose subsequences
$a_{n(1)-1},a_{n(2)-1},\ldots$

$a_{n(1)-2},a_{n(2)-2},\ldots$
$\ldots$
$a_{1},a_{n(2)-n(1)+1},\ldots$
$a_{n(2)-n(1)},a_{n(3)-n(1)},\ldots$

and so on. Again each of these subsequence can not decay faster than $f$.

Am I doing any mistake?

Thanks for your help!

1 Answers 1

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Let $\{M_n:n\in\Bbb Z\}$ be a partition of $\Bbb N$ into infinite sets. For each $n\in\Bbb Z$ let $M_n=\{m(n,k):k\in\Bbb N\}$, where $m(n,0). Let $f(x)=2^{-x}$. For $n\in\Bbb Z$ and $k\in\Bbb N$ let

$a_{m(n,k)}=\frac1{2^{n+m(n,k)}}\;,$

so that

$\frac{a_{m(n,k)}}{f\big(m(n,k)\big)}=\frac{1/2^{n+m(n,k)}}{1/2^{m(n,k)}}=\frac1{2^n}\;.$

Then for each $n\in\Bbb Z$ the subsequence $\langle a_{m(n,k)}:k\in\Bbb N\rangle$ satisfies $a_{m(n,k)}>Cf\big(m(n,k)\big)$ for each $C<2^{-n}$, but clearly

$\liminf_m\frac{a_m}{f(m)}=0\;.$

Thus, there is no $C>0$ such that $a_m>Cf(m)$ for all $m\in\Bbb N$.

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    @Lenava: My $C$ doesn’t actually depend on $n$. Given $C$, choose an appropriate $n$. However, I did overlook the fact that you were interested in arbitrarily **large** values of $C$, not just arbitrarily small ones, so my example handles only values of $C$ less than $1$. I’ve now corrected this. Given C>0, pick $n\in\Bbb Z$ so that 2^{-n}>C, and consider the sequence $\langle a_{m(n,k)}:k\in\Bbb N\rangle$.2012-08-10