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I am investigating a strictly decreasing sequence $(a_i)_{i=0}^\infty$ in $(0, 1)$, with $\lim_{i\to\infty}a_i=0$, such that there exist constants $K>1$ and $m\in\mathbb{N}$ such that $\frac{a_{i-1}^m}{K} \leq a_i \leq K a_{i-1}^m$ for all $i$. Even though $K>1$, is it of the right lines to conclude that $a_i \sim \alpha^{m^i}$ for some constant $0<\alpha<1$?

Thanks, DW

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    @abatkai: It might be hard to see, but the exponent of $\alpha$ is $m^i$ (i.e., $i$ is in the exponent of the exponent).2012-05-31

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[Edit: Now that the question has been changed a day later, I removed analysis of the old version of the first inequality. Perhaps sometime I will update to fully answer the new question. The following still applies to the second inequality.]

On the other hand, f For the second inequality, $a_i\leq Ka_{i-1}^m\implies a_i\leq K^{1+m+m^2+\cdots+m^{i-1}}a_0^{m^i}\leq \left(K^{2/m}a_0\right)^{m^i}.$ You could apply a similar inequality with each $N\in\mathbb N$ in place of $0$ to get $a_i\leq \left(K^{2/(m^{N+1})}a_N^{1/m^N}\right)^{m^i}.$ By choosing $N$ such that $K^{2/m}a_N<1$, you at least get $\displaystyle{a_i=O\left(\alpha^{m^{i}}\right)}$ with $\alpha=\left(K^{2/m}a_N\right)^{1/m^N}\in(0,1)$.

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    @Derek: OK, I've shown a way to use the upper bound. What can you get from the lower bound?2012-06-01