How are these summations equal? $\sum_{q=2}^{\infty} \sum_{p=-mq}^{mq} \frac{2}{q^n} = \sum_{q=2}^{\infty} (2mq + 1) \frac{2}{q^n} $

Question

$$\sum_{q=2}^{\infty} \sum_{p=-mq}^{mq} \frac{2}{q^n} = \sum_{q=2}^{\infty} (2mq + 1) \frac{2}{q^n} $$

I cannot understand how the jump is being made. Would appreciate any help.

I'm reading through Oxtoby's proof of the Lebesgue measure of Liouville numbers being 0.

Answer 1

Please see if you understand this: $\sum_{q=2}^{\infty} \sum_{p=-mq}^{mq} \frac{2}{q^n} = \sum_{q=2}^{\infty} \frac{2}{q^n} \sum_{p=-mq}^{mq} 1$ while $\sum_{p=-mq}^{mq} 1 = 2mq+1$ just counts the number of integers between $-mq$ and $mq$.