Doubly exponential sequence behaviour from inequality

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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The double inequality cannot hold when $a_i$ becomes very small. –  user31373 May 30 '12 at 22:43
Extending the comment of @LeonidKovalev, what does your conclusion mean? Where is the dependence on $i$ on the right side? –  abatkai May 30 '12 at 23:09
@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). –  Jonas Meyer May 31 '12 at 6:15

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 $\alpha=\left(K^{2/m}a_N\right)^{1/m^N}\in(0,1)$.
Thanks @Jonas and others for alerting me to this mistake. The fraction read $\frac{1}{Ka_{i-1}^m}$ but should've, and has been changed, to read $\frac{a_{i-1}^m}{K}$. –  Derek Jun 1 '12 at 16:30