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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!

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1 Answer 1

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 $m(n,0)<m(n,1)<m(n,2)<\ldots\;$. 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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Thank you Brian! But I think you misunderstood me. Because you chose a sequence $a_{m(n,k)}$ decaying faster than f(m(n,k)). That's why your inequality $a_{m(n,k)}> C f(m(n,k))$ is of course true for $C(n)<2^{-n}$ but i meant a $C>0$ not depending on $n$. –  Lenava Aug 10 '12 at 8:15
    
@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 $\langle a_{m(n,k)}:k\in\Bbb N\rangle$. –  Brian M. Scott Aug 10 '12 at 8:25
    
Thank you Brian. I confused n with k. –  Lenava Aug 20 '12 at 11:23
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