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 $K&gt;1$ 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 $K&gt;1$, is it of the right lines to conclude that $a_i \sim \alpha^{m^i}$ for some constant $0&lt;\alpha&lt;1$?

Thanks,
DW
[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 $K^{2/m}a_N&lt;1$, you at least get $\displaystyle{a_i=O\left(\alpha^{m^{i}}\right)}$ with $\alpha=\left(K^{2/m}a_N\right)^{1/m^N}\in(0,1)$.
[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 $K^{2/m}a_N&lt;1$, you at least get $\displaystyle{a_i=O\left(\alpha^{m^{i}}\right)}$ with $\alpha=\left(K^{2/m}a_N\right)^{1/m^N}\in(0,1)$.