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.
Two sequences with convergent ratio
3
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calculus
real-analysis
sequences-and-series
convergence
1 Answers
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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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0That's right, Thanks! – 2012-06-20
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0No problem, glad to help! – 2012-06-20
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0If my answer is satisfactory, you can "accept" it by clicking the checkmark underneath the number and arrows. – 2012-06-20
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0of course it is! – 2012-06-20