I have to write a proof concerning a grim trigger in an infinite Prisoners' Dilemma game. I can write the utility for both players cooperating infinitely. But suppose at some turn $n$ one player decides to pick the defect strategy. Is it then possible/allowed to state that starting with turn $n+1$ it is a 'new' infinite game? Because that would make my proof pretty much trivial :)
For example, considering one player's utility when both cooperate infinitely is $$\sum_{k=1}^{\infty} \delta^{k-1} \cdot -2 = \frac{1}{1-\delta} \cdot -2$$
Can I now say that the utility for that player when defecting in turn $n$ includes another infinite summation like here? $$\sum_{k=1}^{n-1}\left( \delta^{k-1} \cdot -2\right) + \left(\delta^{n-1} \cdot 0\right) + \left( \frac{1}{1-\delta} \cdot -4\right)$$
Almost. You have an infinite series of defections starting in period $n$. However, you still need to discount the infinite sum of these payoffs to get the present value - after all, the defections only start in the future, not right now. Thus, the sum of discounted payoffs should be $$\sum_{k=1}^{n-1}\left( \delta^{k-1} \cdot -2\right) + \left(\delta^{n-1} \cdot 0\right) + \delta^n\left( \frac{1}{1-\delta} \cdot -4\right).$$