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Let $ (b_{n})$ be a decreasing a sequence such that $0< b_{n}<1$ for all $n\geq 1$, and $b_{n}\to 0$ as $n\to \infty$. Is there any way to find another sequence $(a_{n})$ with $\frac{a_{n}}{b_{n}}$ converges to a nonzero constant, and $\frac{a_{n}}{b^{2}_{n}}$ is bounded.

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1 Answer 1

up vote 3 down vote accepted

By definition, if $\frac{a_n}{b_n}\to L$ for some non-zero $L$, then for any $\epsilon>0$, there exists an $N$ such that for all $n>N$, we have $|\frac{a_n}{b_n}-L|<\epsilon$.

Let's take $\epsilon=|\frac{L}{2}|$, so that $|\frac{a_n}{b_n}-L|<|\frac{L}{2}|$ for all $n>N$. In particular, $|\frac{a_n}{b_n}|>|\frac{L}{2}|$ for all $n>N$. Then $$\bigg|\frac{a_n}{b_n^2}\bigg|>\bigg|\frac{L}{2}\bigg|\cdot\frac{1}{b_n},$$ but $\frac{1}{b_n}\to\infty$ because $b_n\to 0$, so $\frac{a_n}{b_n^2}$ cannot be bounded.

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That's right, Thanks! –  Kristina Jun 20 '12 at 1:40
    
No problem, glad to help! –  Zev Chonoles Jun 20 '12 at 1:40
    
If my answer is satisfactory, you can "accept" it by clicking the checkmark underneath the number and arrows. –  Zev Chonoles Jun 20 '12 at 2:05
    
of course it is! –  Kristina Jun 20 '12 at 2:20

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