I'm trying to follow the proof of Lemma 3 page 2: (https://www.math.ubc.ca/~gerg/teaching/613-Winter2011/LargeSieveBombieriVinogradov.pdf) but have gotten a bit stuck.
I'm not following how we go from (page 3 just after "while the second is").
The $ \lambda_{r} $ are a finite collection of distinct real numbers.
$ \sum_{(s,t), s \neq t} \frac{\bar{z_{s}}z_{t}}{\lambda_{s} - \lambda_{t}} \sum_{r \neq s, t} \Big(\frac{1}{\lambda_{r} - \lambda_{s}} - \frac{1}{\lambda_{r} - \lambda_{t}} \Big) $ to
$ \sum_{(s,t), s \neq t} \frac{\bar{z_{s}} z_{t}}{\lambda_{s} - \lambda_{t}} \Big[\sum_{r \neq s} \frac{1}{\lambda_{r} - \lambda_{s}} - \sum_{r \neq t} \frac{1}{\lambda_{r} - \lambda_{t}} + \frac{2}{\lambda_{s} - \lambda_{t}}\Big] $
I then don't quite see how these two summations in the square brackets cancel each other out. I'm guessing it's some kind of telescoping series but i'm not quite seeing it.
This similar question here didn't help me much (Show that the sums $\sum_{r \neq s} \frac{1}{\lambda_r -\lambda_s} - \sum_{r \neq t} \frac{1}{\lambda_r -\lambda_t}$ cancel out.)
Thanks.
You have $r\ne s$ and the sum $$\newcommand{\la}{\lambda}\sum_{r\ne s,t}\left( \frac1{\la_r-\la_s}-\frac1{\la_r-\la_t}\right).$$ Thus is \begin{align} \sum_{r\ne s,t}\frac1{\la_r-\la_s}-\sum_{r\ne s,t}\frac1{\la_r-\la_t} &=\sum_{r\ne s}\frac1{\la_r-\la_s}-\frac1{\la_t-\la_s} -\sum_{r\ne t}\frac1{\la_r-\la_t}+\frac1{\la_s-\la_t}\\ &=\sum_{r\ne s}\frac1{\la_r-\la_s} -\sum_{r\ne t}\frac1{\la_r-\la_t}+\frac2{\la_s-\la_t}. \end{align}