The function is defined as: $$f(t) = \sum_{n=-\infty}^{+\infty} \frac{1-e^{i\gamma^n \,t}}{\gamma^{(2-D)n}}$$
I want to calculate $|f(t+\tau)-f(t)|$. I wrote out the steps I saw somewhere, I just want to make sure that the error that I am seeing is really there. So is this correct or not?
\begin{align}|f(t+\tau)-f(t)| &= \left|\sum_{n=-\infty}^{+\infty} \left[\frac{e^{i\gamma^n\,t}-e^{i\gamma^n(t+\tau)}}{\gamma^{(2-D)n}}\right]\right|\\&= \sum_{n=-\infty}^{+\infty} \left|\frac{e^{i\gamma^n\,t}-e^{i\gamma^n(t+\tau)}}{\gamma^{(2-D)n}}\right| \\&=\sum_{n=-\infty}^{+\infty} \frac{e^{i\gamma^n\,t}-e^{i\gamma^n(t+\tau)}}{\gamma^{(2-D)n}} \frac{e^{-i\gamma^n\,t}-e^{-i\gamma^n(t+\tau)}}{\gamma^{(2-D)n}} \\&= \sum_{n=-\infty}^{+\infty} \frac{1 - e^{-i\gamma^n\,\tau}-e^{i\gamma^n\,\tau}+1}{\gamma^{2(2-D)n}} \\&= \sum_{n=-\infty}^{+\infty} \frac{2 -2\cos{\gamma^n\,\tau}}{\gamma^{2(2-D)n}}\end{align}
Your second equality is False. It is actually an inequality (triangular inequality)