Sum of a dyadic sequence satisfying $\delta\omega_\lambda \leq \omega_{2\lambda}\leq \kappa\omega_\lambda$ with $1<\delta<\kappa$ I was reading the proof of a lemma and I found something that seemed to be a "trivial inequality" but I haven't been able to prove it, so I was wondering if anyone has a good answer for it.
Lets fix two real numbers $1<\delta<\kappa$. Now, consider a sequence $(\omega_{\lambda})_{\lambda}\subset\mathbb{R}_+$ with $\lambda$ running over the integer dyadic numbers and where $\mathbb{R}_+=(0,+\infty)$. Suposse that $(\omega_\lambda)_\lambda$ satisfies the following inequalities: $$
\delta\omega_\lambda \leq \omega_{2\lambda}\leq \kappa\omega_\lambda.
$$
Then, there exists a constant $c>0$ such that $$
\sum_{\mu\geq \lambda/8}\omega_\mu^{-1}\leq c\omega_{\lambda}^{-1}.
$$
Here, notation $\sum_\lambda \omega_\lambda$ means $\sum_{n=0}^{+\infty}\omega_{2^n}$. I think it should be a silly trick using power series but I am getting confuse since I am not very used to dyadic stuffs.
 A: As Steven Stadnicki suggested in the comments, if 'integer dyadics' means numbers of the form $2^n$, we can rewrite the problem with $\mu_i := \omega_{2^i}$. The assumption is now :
$$\forall i \geq 0, \quad \delta \mu_i \leq \mu_{i+1} \leq \kappa \mu_i$$
We want to prove that there exists $c > 0$ such that :
$$\forall n \geq 0,\quad \sum_{i \geq ~\max(n - 3, 0) } \frac1{\mu_i} \leq c \frac1{\mu_n}$$
First, we can suppose $n \geq 3$ because for $0 \leq n \leq 2$ the inequality is true with $c_1 = \mu_2 \times (\sum_{i \geq 0 } \frac1{\mu_i})$.
Heuristically, $\mu_i$ should behave as a geometric sequence. Remember that, to compute the sum $\sum_{i \geq n} q^i$ (with $0 < q < 1$) we use a telescopic sum :
$$\sum_{i \geq n} q^i (1 - q) = \sum_{i \geq n} q^i - q^{i+1} =  q^n $$
In fact, we can do the same with our problem :
$$\sum_{i \geq n-3} \frac1{\mu_i} (1 - \frac1{\delta}) = \sum_{i \geq n-3} \frac1{\mu_i} - \frac1{\mu_i \delta} \leq  \sum_{i \geq n-3} \frac1{\mu_i} - \frac1{\mu_{i+1}} = \frac1{\mu_{n-3}} \leq \kappa^3 \frac1{\mu_{n}} $$
We used the assumption in the two inequalities.
We conclude by taking $c = \max(c_1, \frac{\kappa^3}{(1 - \frac1{\delta})})$
