How to show this inequality in the law of iterated logarithm

Question

Let $\psi(t) = \sqrt{2t\log\log t}$ and $q>4$. Then we have for large $k$,

$$ \psi\left(q^k - q^{k-1}\right) \geq \psi\left(q^k\right) \left(1-\frac{1}{q}\right). $$

To prove this, I can tell by definition it is equivalent to show that

$$ \sqrt{2q^k\left(1 - \frac{1}{q}\right)\log\log q^k\left(1 - \frac{1}{q}\right)} \geq \sqrt{2 q^k \log\log q^k} \left(1-\frac{1}{q}\right) $$

then

$$ \sqrt{\log\log q^k\left(1 - \frac{1}{q}\right)} \geq \sqrt{\left(1 - \frac{1}{q}\right)\log\log q^k} $$

then

$$ \log\log q^k\left(1 - \frac{1}{q}\right) \geq \left(1-\frac{1}{q}\right)\log\log q^k. $$

How to proceed from this, please?

Answer 1

By Hôpital's theorem, we have

$$\lim_{k \to \infty} \frac{\log (a_k+c)}{\log a_k} = 1$$

for any sequence $(a_k)$ such that $a_k \to \infty$ and $c \in \mathbb{R}$. Therefore

$$\frac{ \log \log \left[ q^k \left(1- \frac{1}{q} \right) \right]}{\log \log q^k} = \frac{\log \left(k \log q + \log\left(1- \frac{1}{q} \right) \right)}{\log (k \log q)}$$

shows that

$$\lim_{k \to \infty} \frac{ \log \log \left[ q^k \left(1- \frac{1}{q} \right) \right]}{\log \log q^k}=1.$$

In particular, we can choose $K$ such that

$$ \frac{ \log \log \left[ q^k \left(1- \frac{1}{q} \right) \right]}{\log \log q^k} \geq \left(1- \frac{1}{q}\right)$$

for all $k \geq K$.