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Given the sequences $a_{n},b_{n},c_{n}$ all subsets of the interval $(0,1)$, and all converges to 0. Also we have $a_{n}\leq b_{n}^{\alpha}$, and $a_{n}\leq c_{n}$, for all $n=1,2,3,...$, for some $\alpha \geq 1$ (real number).

Now, for a given $n$, we could have $$a_{n}\leq b_{n}^{\alpha} \leq c_{n}$$ or we could have $$a_{n} \leq c_{n} \leq b_{n}^{\alpha}$$

Suppose that we have the choice to pick any $\alpha\geq 1$. Is it possible to pick an $\alpha$ large enough so that the first case is true for all $n$ (or at least for a subsequence of $n$)?.

Edit: "for some $\alpha$" instead of for all.


I'm trying the following:

If there is $n_{1}, n_{2}, n_{3},...$ such that $b_{n_{i}}^{\alpha} > c_{n_{i}} $, for $i=1,2,3,.. $ ,then we can choose $\alpha_{1}, \alpha_{2}, \alpha_{3},..$ so that $ b_{n_{i}}^{\alpha_{i}} \leq c_{n_{i}}$ for all $i=1,2,3,.. $.

Now, let $ \alpha=\sup \alpha_{i}$ then $ b_{n_{i}}^{\alpha} \leq c_{n_{i}}$ for all $i=1,2,3,.. $. But the problem in this method is that the "sup" could be $\infty$!!

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Wait... "for all $\alpha \ge 1$"? Surely that means that $a_n = 0$ for all $n$, since $0 < b_n < 1$? –  dbaupp Jun 10 '12 at 3:17
    
I meant we have the choice on $\alpha$. –  Alpha Jun 10 '12 at 3:20
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1 Answer

No, not for arbitrary sequences $a_n,b_n,c_n$.

Consider the case when $a_n = c_n = A 2^{-n}$ for a constant $A$ (to allow the condition $a_n \le b_n^\alpha$ to hold) and $b_n = \frac{1}{n+1}$. For the first case to hold, we must have an $\alpha$ such that $$ \left(n+1\right)^{-\alpha} = A 2^{-n}$$ for all $n$, which is impossible.


If the inequalities are strict, so $a_n < c_n$, then the above proof can be adjusted by choosing $a_n = A 2^{-n}$ and $c_n = (A + \epsilon) 2^{-n}$ for some small $\epsilon > 0$ and $b_n$ as before, then we need to find an $\alpha$ such that $$ A 2^{-n} \le \left(n+1\right)^{-\alpha} \le \left(A+\epsilon\right) 2^{-n}$$ for all $n$, which is again impossible.

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That's right. What if $a_{n}< c_{n}$ (strictly less than)? –  Alpha Jun 10 '12 at 8:14
    
@Alpha, I've added that to my answer. (If this answer is useful, you should vote it up, this section of the FAQ has more information about this process towards then end :) ) –  dbaupp Jun 10 '12 at 8:30
    
Okay! Let me ask another question: In general, for any $a,b,c\in (0,1)$, if $a<c<b$, can we find an $\alpha$ such that $a<b^{\alpha}<c$? –  Alpha Jun 10 '12 at 8:53
    
Yes. $\alpha = \log_b \frac{a+c}{2}$. –  dbaupp Jun 10 '12 at 9:06
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