I'm working my way through this paper (https://www.math.ubc.ca/~gerg/teaching/613-Winter2011/LargeSieveBombieriVinogradov.pdf) and am not quite seeing how one goes from (Lemma 3 page 3):
Note: $\lambda_{r}$'s are positive distinct real numbers and the z denote complex numbers.
$\sum_{s} \sum_{t \neq s} \frac{|z_{s}|^{2} + |z_{t}|^{2}}{(\lambda_{s} - \lambda_{t})^{2}} = 2 \sum_{s} |z_{s}|^{2} \sum_{t \neq s} \frac{1}{(\lambda_{s} - \lambda_{t})^{2}} $
The paper states that it is because it is due to the pair $(i,j)$ occuring once as $ s= i, t =j$ and once as $s=j, t=i$
I've tried plugging these in these values into the first sum but i don't see how the result follows.
Thank you.
Does this seem more clear? $$\sum_s \sum_{t \neq s} \frac{|z_s|^2 + |z_t|^2}{(\lambda_s - \lambda_t)^2} = \sum_{(s,t)\\s \neq t} \frac{|z_s|^2 + |z_t|^2}{(\lambda_s - \lambda_t)^2}$$ $$= \sum_{(s,t)\\s \neq t} \frac{|z_s|^2}{(\lambda_s - \lambda_t)^2} + \sum_{(s,t)\\s \neq t} \frac{|z_t|^2}{(\lambda_s - \lambda_t)^2} = 2 \sum_{(s,t)\\s \neq t} \frac{|z_s|^2}{(\lambda_s - \lambda_t)^2}$$ $$= 2 \sum_s \sum_{t \neq s} \frac{|z_s|^2}{(\lambda_s - \lambda_t)^2} = 2 \sum_s |z_s|^2 \sum_{t \neq s} \frac{1}{(\lambda_s - \lambda_t)^2}.$$