Apostol's IANT Section 11.6 on the half-plane of convergence presents and proves a Lemma 2.
Question: I can't understand the simplifications he does at the last step. The image below highlights the step.
Notes: My own attempts at explaining is to use
$$ \left|b^{\sigma_{0}-\sigma}-a^{\sigma_{0}-\sigma}\right|=\left|a^{\sigma_{0}-\sigma}-b^{\sigma_{0}-\sigma}\right|<a^{\sigma_{0}-\sigma}<2a^{\sigma_{0}-\sigma}$$
Is this correct?
It is known that $a<b$, so we have
$$ |b^{\sigma_0-\sigma}-a^{\sigma_0-\sigma}|=a^{\sigma_0-\sigma}-b^{\sigma_0-\sigma} $$
Since it is also known that $b^{\sigma_0-\sigma}>0$, we obtain
$$ |b^{\sigma_0-\sigma}-a^{\sigma_0-\sigma}|<a^{\sigma_0-\sigma} $$